MEI C2 Study Resources Core2 Differentiation 1-2

MEI Core 2 Differentiation Section 1: Introduction to differentiation Study Plan Background Finding the gradient of a curve at a specific point was a chore at GCSE. You had to draw a tangent to the curve at the point in question as accurately as you could, then use its gradient to estimate the gradient of the curve at the point. This section introduces a method of finding gradients of curves by calculation, using a very powerful mathematical technique called differentiation.

Detailed work plan 1. Start by reading pages 191 – 193, which introduces the ideas in this chapter. 2. To illustrate these ideas, look at the Flash resources The limit of the gradient of a chord on y = x², The limit of the gradient of a chord on y = x³, Gradient of a curve at a point on y = x², and Gradient of a curve at a point on y = x³. 3. This work is now formalised. The concepts behind this can appear quite daunting at first so you will need to read pages 194 to 195 quite carefully and then work through activity 8.3. Remind yourself of the binomial theorem from chapter 3 and then read example 8.1. Another example like this one is given in the Notes and Examples (Example 5). The alternative notation shown at the bottom of page 197 is quite common in questions and texts, so it is worthwhile making sure that you are equally comfortable with this as with the notation earlier in the section. 4. Exercise 8A Work through the whole of this exercise following example 8.1. 5. Read pages 197 – 200. This is the usual process that you will require when using differentiation. Make a note of the general results given at the bottom of page 199 and study examples 8.3 and 8.4. Further examples are given in the Notes and Examples. Be precise with how you set out work. Don’t put equal signs where they don’t belong. dy e.g. y = x2 = 2x is nonsense. You should write y = x2, = 2x. dx 6. You can see some examples using the Flash resources Basic differentiation: individual powers of x and Basic differentiation: sums of powers of x.

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MEI C2 Differentiation Section 1 Study plan 7. Exercise 8B It is worth doing all of the questions in this exercise, as practice will ensure comfortable familiarity with the process. The last six parts of question 1 have worked solutions on the website for you to download. You can also try the Differentiation dominoes activity. 8. For extra practice try the interactive questions Basic differentiation. 9. Read examples 8.5 and 8.6 (pages 201 – 202). If you are unsure about solving polynomial equations or sketching curves go back over the work that you did in Chapter 3. You will need these skills in the following and future exercises. 10. Exercise 8C Try the following questions. Those with * beside them have worked solutions for you to download. 1, 2, 3, 4*, 5*, 7, 8*, 10, 12* Knowing how to find the equation of a tangent or a normal to a curve is a useful skill in many areas of mathematics and science. Now that you know how to find the gradient of a curve by differentiation, and hence the gradient of the tangent at that point, you can use the techniques that you learnt in chapter 2 to get the equation of the tangent/normal. 11. Revise the different forms that the equation of a straight line can take and the formulae that can be used to find them. Read examples 8.7 and 8.8 on pages 206 and 207, and Examples 6 and 7 in the Notes and Examples, and then try the questions below. As usual, those with an asterisk have worked solutions available on the website. 12. You can see some further examples using the Flash resources Tangent at a point on a quadratic curve and Normal at a point on a quadratic curve. You may also find the Mathcentre video Tangents and normals useful. Note that this video may contain examples which use some differentiation techniques that you have not yet met. 13. Exercise 8D 1, 3*, 4, 6, 7, 8*, 10*, 11*, 13 14. For extra practice, try the interactive questions Finding tangents and normals and Finding points on curves, given gradients.

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MEI Core 2 Differentiation Section 1: Introduction to differentiation Notes and Examples These notes include sub-sections on:  What is differentiation?  Investigating gradients  Rules for finding derivatives  Differentiating from first principles  Finding tangents and normals to curves

What is differentiation? In this section, you will be studying the relationship between the position of a point on a curve and the gradient of the curve. Straight lines are, by definition, lines of constant gradient. Curves, on the other hand, have varying gradient – the gradient depends on whereabouts you are on the curve. Differentiation is the process of finding the gradient at any point on a curve from the equation of the curve. Differentiation, together with its reverse process, called integration, form the branch of mathematics called calculus. The discovery of calculus (Made in the 17th century by Isaac Newton in England and, independently, by Gottfried von Liebnitz in Germany) was one of the most significant advances in the history of mathematics and science, and was crucial to unlocking the mathematical basis of our planetary system. Differentiation is the process of finding the gradient function, or derivative, or derived function. Given an equation for y in terms of x, the gradient function or dy derivative is written , and gives the gradient of the curve in terms of x. dx

Investigating gradients You can look at a demonstration of how the gradients of chords approach the gradient of a tangent using the Flash resource The limit of the gradient of a chord on y = x². This looks at the example on page 193. You can see another example using the Flash resource The limit of the gradient of a chord on y = x³. To investigate the pattern in the value of the gradient at different points on a curve, use the Flash resources Gradient of a curve at a point on y = x² and Gradient of a curve at a point on y = x³.

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MEI C2 Differentiation Section 1 Notes and Examples Rules for finding derivatives. If y is a polynomial function (made up of powers of x), the following rules will dy enable you to find the derivative : dx n n1  The derivative of x is nx ,  The derivative of kxn is knxn1,  The derivative of a constant is zero.  The derivative of a sum (or difference) is the sum (or difference) of the derivatives

Example 1 Differentiate y = 2x3  5x2 + 4 Solution The derivative of 2x3 is 2 × 3x2 = 6x2 The derivative of 5x2 is 5 × 2x = 10x The derivative of 4 is 0. So

dy = 6x2  10x. dx

You can see further examples using the Flash resources Basic differentiation: individual powers of x and Basic differentiation: sums of powers of x. For practice in questions like the one above, try the interactive questions Basic differentiation. For additional practice, try the Differentiation Dominoes activity. Cut out the dominoes and match each expression (on the blue right side of a domino) with its derivative (on the yellow left side of a domino). The dominoes should form a single loop.

The next example involves an expression which is the product of two functions. You cannot differentiate this by differentiating each function separately and then multiplying the results, i.e. the derivative of a product of two functions is not the product of the derivatives! So with examples involving brackets, you will need to multiply out the brackets first. (There is a rule for differentiating the product of two functions, but you do not need to know this yet.) Example 2 (i) Find the derivative of y = (x  2)(x2 + 1). (ii) Hence find the gradient of the curve at the point (2, 0) (iii) Find the coordinates of the points where the gradient is zero.

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MEI C2 Differentiation Section 1 Notes and Examples Solution (i) y = (x  2)(x2 + 1) = dy dx

 (ii)

x3

= 3x

2

 2x2

+ x

 4x

+ 1

 2

Substituting x = 2 into the gradient function,

dy  3  22  4  2  1 dx 5 so the gradient of the curve at (2, 0) is 5. (iii)

The gradient of the curve is zero when dy = 0.  3x  4x + 1 = 0  (3x  1)(x  1) = 0 1  x = 3 and x = 1.

dx

2

When x =

1 3

,y=

Now calculate the y coordinates for these values of x.

 13  2 91  1   53  109   2750

When x = 1, y = (1  2)(12 + 1) = 2. So the points on the curve with gradient zero are (1, 2) and

 13 ,  5027 

The points where the gradient is zero are called the turning points or stationary points of the curve. You will look at such points in more detail in Section 2.

The next example involves the quotient of two functions (i.e. one function divided by another). As with products the derivative of a quotient is not the quotient of the derivatives. You need to divide the fraction first.

Example 3

3x 2  4 x 4 Differentiate . 2x Solution 3x 2  4 x 4 3x 2 4 x 4   y= = 2x 2x 2x 

3 2

x  2 x3

dy 3 = 2  6x 2 dx

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MEI C2 Differentiation Section 1 Notes and Examples

Example 4 Sketch the graph of the gradient function of the curve shown below. y 2 2

-2

-1 -2

0 -1

1 0

2 1

2

Gradient zero at these points, so derivative is zero

-3 -3

Solution

dy dx2

Gradient positive So derivative positive

-2

-1

Gradient negative So derivative negative

0

1

2

-3

Differentiating from first principles dy for the derivative of a function comes from differentiation dx from first principles. To find the gradient function of a function f, you find the gradient of the chord which joins the point (x, f(x)) to another point (x + x, f(x + x)), where x is very small. When you have simplified this as much as possible, you then let x = 0, and the chord becomes a tangent to the graph. Example 8.1 in the textbook illustrates this method for the function y  x3 . Here is a further example. The notation

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MEI C2 Differentiation Section 1 Notes and Examples Example 5 Differentiate y = x3  2x from first principles. Solution y + y = (x + x)3  2(x + x) and y = x3  2x Subtracting gives Dividing by x:

Now letting x 0,

y = (x + x)3  2(x + x)  x3 + 2x

δy ( x  δx)3  2( x  δx)  x3  2 x = δx δx 3 2 x  3x δx  3x(δx)2  (δx)3  2 x  2δx  x3  2 x = δx 2 2 3x δx  3x(δx)  (δx)3  2δx = δx 2 = 3x  3x(δx)  (δx)2  2 δy dy  = 3x2  2. δx dx

Notice that this is the same answer you would obtain by applying the rules to dy find . dx By differentiating from first principles you can see that the rules do indeed give the correct gradient functions (derivatives).

Finding Tangents and Normals to Curves The gradient of a tangent to a curve at a particular point is the same as the gradient of the curve at that point. So to find the equation of a tangent to a curve, you first need to find the gradient m of the curve via differentiation. You can then substitute m and the coordinates (x1, y1) of the point on the curve into the formula: y  y1 = m(x  x1). You can look at examples of tangents to a quadratic curve using the Flash resource Tangent at a point on a quadratic curve.

Example 6 Find the equation of the tangent to the curve y = 2x3  3x at the point with xcoordinate 1.

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MEI C2 Differentiation Section 1 Notes and Examples Solution y = 2x3 3x 

To find the gradient of the tangent, first differentiate the equation of the curve.

dy = 6x2  3. dx

dy   6 12   3  3 dx The gradient of the tangent is therefore 3. Find the y-coordinate of the point At the point with x-coordinate 1,

where x = 1 by substituting into the equation of the curve.

y = 2x  3x x = 1  y   2 13    3 1  2  3  1 3

y  y1  m( x  x1 ) y  (1)  3( x  1) y  1  3x  3 y  3x  4 This is the required equation of the tangent.

Use the formula for the equation of a line with m = 3, x1 = 1 and y1 = 1

The normal to a curve is the line perpendicular to the tangent. Remember that the gradient of a line perpendicular to a line with gradient m is m, where

m  

1 . m

You can look at examples of normals to a quadratic curve using the Flash resource Normal at a point on a quadratic curve.

Example 7 Show that the normal to the curve y = 2x2  3x at the point (1, 1) passes through the origin. Solution dy y = 2x  3x  = 4x  3. dx dy   4 1  3  1 At the point with xcoordinate 1, dx Gradient of the tangent = 1 2

First, find the gradient of the tangent by differentiating y.

1 so gradient of the normal =   1 1 y  y1  m( x  x1 ) y  (1)  1( x  1) y 1  x 1 y  x This line passes through the origin.

Use the formula for the equation of a line with m = -1, x1 = 1 and y1 = 1

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MEI C2 Differentiation Section 1 Notes and Examples You may find the Mathcentre video Tangents and normals useful. Note that this video may contain examples which use some differentiation techniques that you have not yet met.

For practice in questions like Examples 6 and 7, try the interactive questions Finding tangents and normals.

Example 8 A curve has equation y  x3  x 2  x  2 . (i) Find the x-coordinates of the points on the curve with gradient 6. (ii) Find the x-coordinates of the points on the curve for which the gradient of the normal is  12 . Solution

dy  3x 2  2 x  1 dx Gradient = 6  3x 2  2 x  1  6

y  x3  x 2  x  2  (i)

 3x 2  2 x  5  0  (3 x  5)( x  1)  0  x  53 or x  1 (ii)

Gradient of normal =  12  gradient of curve = 2 Gradient = 2

 3x 2  2 x  1  2  3x 2  2 x  1  0  (3x  1)( x  1)  0  x   13 or x  1

For practice in questions like the one above, try the interactive questions Finding points on curves, given gradients.

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Core 2 Differentiation Section 1: Introduction to differentiation Crucial points 1. Remember to multiply out before differentiating a product Sums and differences of terms are easy to differentiate – just differentiate each part separately, and put the derivatives of the parts together. However, you cannot treat products in the same way. There is a product rule for differentiating, but this is not part of the AS syllabus. At this stage, you will need to multiply out the product first, then differentiate the result term-by-term. Example:

Differentiate y = 3x²(x² − 2).

²

Wrong:

3

Right:

y = 3 x 2 ( x 2 − 2) dy = 6 x × 2 x = 12 x 2 dx

²

y = 3x 2 ( x 2 − 2) y = 3x 4 − 6 x 2 dy = 12 x 3 − 12 x dx

3

2. Rewrite quotients before differentiating As with products, quotients have a special rule for differentiating them. At this stage, you can only differentiate these by dividing them out into separate terms. x3 + 2 x 4 . Example: Differentiate y = 4x2

²

Wrong

3

Right

dy 3 x 2 + 8 x 3 = dx 8x

²

x3 2 x 4 1 + = 4 x + 12 x 2 4 x2 4 x2 dy 1 1 = 4 + 2 × 2 x = 14 + x dx

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Core 2 3. Make sure you differentiate with respect to the correct variable Example: Differentiate x2 + a2 with respect to x, where a is a constant.

²

Wrong

dy = 2 x + 2a dx

3

Right

dy = 2x dx

The a² term is a constant, so its derivative is zero

²

3

4. Use notation correctly Example: Differentiate v = t2 + 2t.

The expression you are differentiating has variables v and t, not y and x, so you

²

Wrong

dy = 2t + 2 . dx

3

Right

dv d 2 (t + 2t ) = 2t + 2 . = 2t + 2 , or dt dt

²

are finding

dv dy , not . dt dx

3

5. When calculating a gradient or tangent to a curve, make sure you get the coordinates the right way round Example: Find the gradient of the curve y = x2 + 2x − 3 where it crosses the x−axis.

²

dy = 2x + 2 . dx

Wrong

When x = 0,

3

dy = (2 × 0) + 2 = 2 . dx

²

dy = 2x + 2 . dx Curve crosses x axis when y = 0, ⇒ x2 + 2 x − 3 = 0

Right

⇒ ( x − 1)( x + 3) = 0 ⇒ x = 1 or x = −3 dy When x = 1, = (2 × 1) + 2 = 4 dx dy When x = −3, = (2 × −3) + 2 = −4 dx 7. Don’t mix up

dy δy and dx δx

In differentiating from first principles, don’t get

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dy δy and mixed up. δy dx δx

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Core 2 and δx represent small changes in x and y – it is only in the limit as we let δx approach zero that we use dy and dx.

8. Be careful with the algebra in first principles differentiation Remember that δx is a single algebraic variable: we can’t treat the ‘δ’ and the ‘x’ as distinct variables

²

Wrong

( x + δx )

2

= x 2 + 2 xδx + δ 2 x 2

²

3

Right

( x + δx )

2

= x 2 + 2 xδx + (δx )

3

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MEI Core 2 Differentiation Section 1: Introduction to differentiation Exercise 1. Differentiate with respect to x: (i) 2x + 1 (ii) x3 − 5x

(iii) x(x + 2).

2. Given that y = 12x − x3, (i) Find the gradient of the curve at the origin. (ii) Find the coordinates of the two points where the gradient is zero. 3. Find the equation of the tangent to the curve y = x4 − x + 1 at the point with x−coordinate 1. 4. Show that the equation of the normal to the curve y = x2 − x at the point (3, 6) is x + 5y = 33. Find the coordinates of the point where the normal meets the x−axis. 5. The displacement s metres of a particle from a point O after t seconds is given by ds the equation s = t3 − 3t2 − 9t. Find the velocity v ( = ) in terms of t, and hence dt find the time at which the particle is stationary (i.e. the velocity is zero).

6. Copy the curve graphed below, and sketch the shape of the derivative on the same axes. y x −2

−1

1

2

−2

−4

dy . Hence find the x−coordinates of the two points dx on the curve where the gradient is 4.

7. Given that y = x3 + 2x2, find

8. (i) Show that the point (1, 2) lies on both the curves y = 2x3 and y = 3x2 − 1. (ii) Show that the curves have the same gradient at this point. (iii) What do these results this tell you about the two curves? 9. A curve has equation y = ax3 + bx, where a and b are constants. At the point where x = 1, the y −coordinate is 8 and the gradient is 12. Find a and b. 10. Show that the tangent to the curve y = x3 + x + 2 at the point P with x−coordinate 1 passes through the origin, and find the equation of the normal at this point. Given that the normal cuts the x −axis at the point Q, find the area of triangle OPQ.

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MEI Core 2 Differentiation Section 1: Introduction to differentiation Solutions to Exercise 1.

(i) f( x ) = 2 x + 1 f ′( x ) = 2 (ii) f( x ) = x 3 − 5 x f ′( x ) = 3 x 2 − 5 (iii) f( x ) = x( x + 2) = x 2 + 2 x f ′( x ) = 2 x + 2

2. (i) y = 12 x − x 3

dy dx

= 12 − 3 x 2 dy

= 12 dx The gradient of the curve at the origin is 12.

When x = 0,

(ii) When gradient is zero, 12 − 3 x 2 = 0 4− x2 = 0 (2 + x )(2 − x ) = 0 x = −2 or x = 2 When x = -2, y = 12 × −2 − ( −2)3 = −24 + 8 = −16 When x = 2, y = 12 × 2 − 2 3 = 24 − 8 = 16 The gradient is zero at (-2, -16) and (2, 16). 3. y = x 4 − x + 1 dy = 4x 3 − 1 dx dy When x = 1, = 4 × 13 − 1 = 4 − 1 = 3 dx When x = 1, y = 1 4 − 1 + 1 = 1 The tangent is the straight line with gradient 3 passing through (1, 1).

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MEI C2 Differentiation Section 1 Exercise solutions Equation of tangent is y − 1 = 3( x − 1) y − 1 = 3x − 3 y = 3x − 2

4. y = x 2 − x dy = 2x − 1 dx

dy

= 2×3− 1 = 5 dx Gradient of tangent = 5, so gradient of normal = − 51 . The normal is the straight line with gradient − 51 passing through (3, 6). Equation of normal is y − 6 = − 51 ( x − 3)

When x = 3,

5( y − 6) = −( x − 3) 5 y − 30 = − x + 3 5 y + x = 33 Where the normal meets the x-axis, y = 0 so x = 33. The normal meets the x-axis at (33, 0). 5. s = t 3 − 3t 2 − 9t ds = 3t 2 − 6t − 9 dt When particle is stationary, 3t 2 − 6t − 9 = 0

v =

t 2 − 2t − 3 = 0 (t − 3)(t + 1) = 0 t = 3 or t = −1

Since time must be positive, t = 3. The particle is stationary after 3 seconds. 6.

gradient graph 4

y

3 2 1 −2

−1

−1

x 1

2

−2 −3 −4

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MEI C2 Differentiation Section 1 Exercise solutions 7. y = x 3 + 2 x 2

dy

= 3x 2 + 4x

dx When gradient is 4, 3 x 2 + 4 x = 4 3x 2 + 4x − 4 = 0 (3 x − 2)( x + 2) = 0

x=

2 3

or x = −2

8. (i) When x = 1, y = 2 x 3 = 2 × 1 3 = 2 When x = 1, y = 3 x 2 − 1 = 3 × 1 2 − 1 = 2 so the point (1, 2) lies on both curves. (ii) y = 2 x 3

dy

= 6x dx When x = 1, gradient = 6 × 1 = 6 y = 3x 2 − 1 dy

= 6x dx When x = 1, gradient = 6 × 1 = 6 so the curves have the same gradient at this point.

(iii) The two curves touch each other at (1, 2). 9. y = ax 3 + bx When x = 1, y = a + b ⇒ a + b = 8 dy = 3ax 2 + b dx When x = 1, gradient = 3a + b ⇒ 3a + b = 12

3a + b = 12

a +b = 8 Subtracting:

2a = 4

a = 2, b = 6

10. y = x 3 + x + 2 dy = 3x 2 + 1 dx

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MEI C2 Differentiation Section 1 Exercise solutions When x = 1,

dy

= 3 × 12 + 1 = 4

dx When x = 1, y = 1 3 + 1 + 2 = 4 The tangent has gradient 4 and passes through the point (1, 4). Equation of tangent is y − 4 = 4( x − 1) y − 4 = 4x − 4 y = 4x So the tangent passes through the origin. Gradient of normal = − 41 Equation of normal is y − 4 = − 41 ( x − 1) 4( y − 4) = −( x − 1) 4 y − 16 = − x + 1 4 y + x = 17 When y = 0, x = 17, so Q is (17, 0).

Area of triangle = 21 × base × height = 21 × 17 × 4 = 34

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Differentiation dominoes Cut out the dominoes along the bold lines and match them end-to-end so that equivalent expressions are adjacent. The dominoes will form a closed loop.

dy  26 x12 dx

y  x2

dy  12 dx

y  4  3x 4

dy  55 x10 dx

y  7x  4

dy 0 dx

y  4x

dy  2x dx

y  4x

dy  12 x3 dx

yx

dy  12 x 2 dx

y  2x  3

dy 2 dx

y  x4

dy  30 x dx

y  x  x  1

dy  2x 1 dx

y  5 x11

dy  63x8 dx

y  2 x 2 1  4 x 

dy 4 dx

y  2 x6

dy 1 dx

y  2 x  4x2

dy  2  8x dx

y  15 x 2

dy  9  2x dx

y  2 x13

dy  4x  3 dx

y  24

dy  4 x  24 x 2 dx

y  9x  6  x2

dy  4 x3 dx

y  2 x 2  3x

dy  12 x5 dx

dy  2 x3  x dx

dy  6 x2 1 dx

y  12 x  9

9

3

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Core 2 Differentiation Section 1: Introduction to differentiation Multiple choice Test 1. The gradient of the curve y = 3x2 − 4 at the point (2, 8) is: (a) 12 (c) 48 (e) I don’t know

(b) 6x (d) 8

2. The derivative of (2x − 1)(x2 − 3) is: (a) 4x (c) 6x2 − 2x − 6

(b) 6x2 − 2x − 6 + c (d)

1 4 1 3 x − x − 3x 2 + 3x + c 4 3

(e) I don’t know 3. The equation of the tangent to the curve y = x3 − x + 3 at the point (1, 3) is: (a) y = 26x − 23 (c) y = 2x (e) I don’t know 3t 3 − 2t 2 , then: 2t 2 dy 3 (a) = dx 2 dx (c) = 1.5 dt (e) I don’t know

(b) y = 2x + 1 (d) y = 2x − 5

4. If x =

dy 1 =1 −t dx 2 dt 3 (d) = dx 2 (b)

5. The normal at the point P to the curve y = 2 − x3 has gradient 3. The x−coordinate of P is: (a) −1 or 1 (c) 1

(b)

1 3

(d) −

1 1 or 3 3

(e) I don’t know

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Core 2 6. The derivative of the equation with this curve

(a)

(b)

(c)

(d)

is

(e) I don’t know.

7. Find the range of values of x for which the gradient of the curve y = x4 − 2x2 is negative: (a) −1 < x < 1 (c) −1 < x < 1 or x > 1 (e) I don’t know.

(b) 0 < x < 1 or x< −1 (d) x > 1 or x < −1

8. The coordinates of the points on the curve y = x 3 − 2x 2 where the gradient is −1 are:

(a) (1, −1) and

( 13 , − 275 )

(c) (1, −1) and ( − 13 ,

5 27

)

(b) (1, −1) and

( 13 , − 1)

(d) (−1, 1) and

( 13 , − 275 )

(e) I don’t know

9. If x =

1+ 4y3 , then: 3

dy = 1 + 4y2 dx dx 1 = + 4y2 (c) dy 3 (e) I don’t know (a)

dy = 4y2 dx dx (d) = 4y2 dy

(b)

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Core 2 10. If, for a particular curve, y = 0 and (a)

dy > 0 when x = 2, then the curve looks like: dx (b) 2

2 (c)

2

(d)

2

(e) I don’t know

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MEI Core 2 Differentiation Section 2: Applications of differentiation Study plan Background In many areas of life we want to know what the maximum value might be or the minimum amount required. This section shows how to identify maximum and minimum values on curves. Being able to do this has direct applications in many scientific and commercial situations. E.g. what is the minimum amount of material needed to package a given article, what would the maximum speed be on a roller coaster of a given shape.

Detailed work plan 1. Read pages 210 to 214 taking careful note of the fact that to identify the nature of a turning point you must look at a point to the left of it first and then at one to the right of it, i.e. always work in the positive direction. There are other ways of identifying max and min but this way is ‘foolproof’ if you remember the hint above. In addition, if you have two turning points close together you must pick points to test that fall on either side of only one of them. If your two test points enclose both turning points, you will get misleading information. Example 1 in the Notes and Examples is another example showing the identification of stationary points, and there is also a PowerPoint presentation for this example. 2. You can test yourself using the Flash resources Stationary points on quadratic curves and Stationary points on cubic curves. 3. Now try the questions below. Exercise 8E 1, 3*, 4, 5, 6, 8*, 11*, 12, 13*, 14, 15* 4. As you saw with question 11 in Exercise 8E it is possible for the gradient of a curve to be zero at a point that is neither a max nor a min. Look carefully at figures 8.24 and 8.25 on page 218 so that you have a clear picture in your mind of a point of inflection. Try the Brand’s Hatch exercise on page 219, visualising how a driver would have to turn his wheel in order to navigate it safely. Then read Example 8.13 on pages 219 – 220. 5. Exercise 8F Try all three questions. There is a worked solution to Q2 on the website. (This question illustrates the note about taking care when choosing which points to test on either side of the turning point.) This is

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MEI C2 Differentiation Section 2 Study plan only one method used to identify stationary points; another is explained below, however it does have its drawbacks. 6. Read the explanations and examples on pages 221 to 225 paying particular attention to the warning note at the top of page 225. This is the drawback mentioned above. You need to be very clear in your mind that just because the first and second derivatives are zero, this does not mean that you have a stationary point of inflection. Think back to the work that you did in chapter 6 and the use of ,, and . This is one of those situations where  is not to be used. There are some additional examples in the Notes and Examples. 7. For additional examples on the second derivative, look at the Flash resource Second derivatives. You may also find the Mathcentre video Maxima and minima useful. (Note that this video may contain examples which use some differentiation techniques that you have not yet met). 8. Exercise 8G Try 3, 4*, 6, 7*, 8, 9*, 10 9. Read pages 227 to 229. This section also gives you some idea of the applications of the work in this chapter. The note at the end of example 8.18 shows how this could link to work done at G.C.S.E. When you were studying distance time graphs, you learnt that the gradient gave you speed; the gradient of speed time graphs gave you acceleration. The work at G.C.S.E. was usually concerned with graphs that came in straight-line sections but if you think about it this is not the most realistic depiction of a journey. Cars do have to speed up and slow down, we don’t always walk at a constant speed etc. We can’t go from 0 to 60km/h instantaneously. This section shows how the work in this chapter can be applied to give more realistic graphs, more realistic information. There is a further example in the Notes and Examples. 10. Exercise 8H Try 2*, 4*, 5, 6*, 8, 9*

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MEI Core 2 Differentiation Section 2: Applications of differentiation Notes and Examples These notes contain sub-sections on:  Turning points  Second derivatives  Maximum and minimum problems

Turning Points Points on a curve where the tangent is horizontal are called stationary points, or turning points. dy At these points, the gradient of the curve is zero, so = 0. dx Stationary points are classified into three different types: Local maximum The gradient is positive to the left, zero at the point, and negative to the right.

gradient positive

gradient zero

gradient negative

gradient negative

gradient zero

gradient positive

Local minimum The gradient is negative to the left, zero at the point, and positive to the right.

Stationary point of inflection The gradient goes from positive to zero to positive

gradient gradient positive zero

gradient positive

or negative to zero to negative.

gradient gradient gradient negative zero negative

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MEI C2 Differentiation Section 2 Notes and Examples To distinguish between these, you can either:  

Test the value of the derivative either side of the stationary point, or Use the second derivative test - see later.

Example 1 Find the stationary points on the curve y = 3x  x3, investigate their nature, and sketch the curve. Step 1: Differentiate the function.

Solution y = 3x  x3 dy = 3  3x2. dx   

Step 2: Solve

3  3x2 = 0 3 = 3x2 x2 = 1 x =  1 or 1

dy

0

dx

When x = 1, y = 3×(1)  (1)3 = 3  (1) = 2; When x = 1, y = 3×1  13 = 2. Step 3: Calculate the ycoordinates for these values of x (called the stationary values).

So the stationary points are (1, 2) and (1, 2).

Step 4: Use a table to investigate the sign of

dy dx

for values of x either side of the stationary values

x dy dx

-2 -9 -ve

-1 0

0 3 +ve

1 0

2 -9 -ve

So ( 1,  2) is a local minimum and (1, 2) is a local maximum The curve crosses the y axis when x = 0. When x = 0, y = 0.

Step 5: Find where the curve cuts the axes

It crosses the xaxis when y = 0. When y = 0, 3x  x3 = 0  x(3  x2) = 0,  x = 0, 3 and  3 .

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MEI C2 Differentiation Section 2 Notes and Examples y (1, 2)

 3

0

x

3

Step 6: Sketch the curve. Make sure your sketch includes the coordinates of the intercepts and the turning points.

(1, 2)

You may like to look at the PowerPoint animation of this example.

For extra practice in finding stationary points, use the Flash resources Stationary points on quadratic curves and Stationary points on cubic curves. For each problem, try to find and identify the stationary points yourself, then check your answer and look at the graph.

Second Derivatives If you differentiate a derivative, you get the second derivative. If you start with an equation for y in terms of x, the first derivative is dee x”) and the second derivative is written

d2 y dx 2

dy dx

(you say: “dee y by

(you say: “dee two y by dee x

squared”)

Example 2 Given that y = 3x  x3, find

d2 y . dx 2

Solution y = 3x  x3 dy  = 3  3x2 dx d2 y  = 6x dx 2

For some more examples, look at the Flash resource Second derivatives. One important application of second derivatives is in the motion of a particle. If you study Mechanics 1 you will learn more about this. If you start with

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MEI C2 Differentiation Section 2 Notes and Examples ds gives the rate dt d2s of change of displacement, or the velocity, and the second derivative 2 dt gives the rate of change of velocity, or acceleration. displacement s as a function of time t, then the first derivative

Example 3 The displacement s metres of a particle from a point P at time t seconds is given by the equation: s = t3  4t. Calculate the velocity and the acceleration of the particle after 2 seconds have elapsed. Solution

s = t3  4t ds  v= = 3t2  4 dt d2s  a = 2 = 6t dt So when t = 2, v = 3 × 22  4 = 8 m s1 a = 6 × 2 = 12 m s2

The second derivative measures the rate of change of the first derivative. We can use this fact to investigate the nature of turning points:

Maximum points dy d2 y If < 0, the gradient function is decreasing. 2 dx dx At a maximum point, the gradient goes from + to 0 to , in other words is decreasing. So

d2 y < 0  the turning point is a maximum. dx 2

Minimum points dy d2 y If > 0, the gradient function is increasing. 2 dx dx At a minimum point, the gradient goes from  to 0 to +, in other words is increasing. So

gradient negative

gradient positive

gradient negative

gradient zero

gradient zero

gradient positive

d2 y >0  the turning point is a minimum. dx 2

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MEI C2 Differentiation Section 2 Notes and Examples Points of inflection You might think that when

d2 y dx 2

= 0, the turning point

is a stationary point of inflection, and it is certainly true that if the stationary point is an inflection, then the value of the second derivative at the point is indeed zero.

gradient gradient positive zero

However, the converse statement is not true: d2 y dy You cannot conclude that if  0 and  0 , the point is a stationary dx 2 dx inflection. In this case, you need to use the table method to investigate the turning point, or consider what you know about the graph of the function. For example, consider y  x 4 .

You may find the Mathcentre video Maxima and minima useful. Note that this video may contain examples which use some differentiation techniques that you have not yet met.

Example 4 Check the nature of the turning points of y = 3x  x3 using the second derivative test. Solution y = 3x  x3 dy  = 3  3x2 dx



dy = 0 when 3  3x2 = 0 dx

x = 1 or  1 When x = 1, y = 2; when x = 1, y = 2 The stationary points are (1, 2) and (-1, -2) d2 y = 6x dx 2 2 When x = 1, d y2 =  6 < 0  maximum

dx 2 When x =  1, d y2 = 6 > 0  minimum. dx

(1, 2) is a maximum point and (-1, -2) is a minimum point.

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gradient positive

MEI C2 Differentiation Section 2 Notes and Examples Maximum and Minimum Problems One important immediate application of differentiation is to problems that involve maximising or minimising a variable quantity.

Example 5 A rectangular sheet of metal of length 1 m and width 1 m has squares cut from each corner. The sides are then folded up to form an open topped box. Find the maximum possible volume of the box. 1 Solution Step 1: Draw a diagram and use x 1 – 2x x

it to help you to formulate the problem mathematically. Call the side length of the squares cut out x. What are the length, width and depth of the box in terms of x?

Length = width = 1 – 2x Depth = x The volume V m3 of the box is given by V = x(1  2x)2. Step 2: Find the maximum volume by differentiating. dV  0 . Before The maximum volume will occur when dx differentiating, expand the brackets.

V  x(1  2 x)2  x 1  4 x  4 x 2   x  4 x 2  4 x3

dV = 1  8x + 12x2 dx

Now put

dV

 0 , and solve for x:

dx

1  8x + 12x2 = 0  (1  2x)(1  6x) = 0  x = 12 or x = 16 . When x  12 , V  12 1  2   12    0 . This must be the minimum. 2

When x  16 , V  16 1  2   16    544  2

2 27

. This must be the maximum.

So the maximum possible volume of the box is

2 27

m3 .

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MEI Core 2 Differentiation Section 2: Applications of differentiation Crucial points (Also see Crucial points in Differentiation 1 – you still need to be looking out for these here!) 1. Take care which side of the stationary point you test the gradient When identifying whether a stationary point is a maximum or minimum by testing the sign of gradient either side of the stationary point, make sure you work from left to right, so you find the gradient at a value of x BEFORE the stationary point first, then at a value of x AFTER the stationary point. Remember: For a maximum, the gradient is positive before the stationary point and negative after it For a minimum the gradient is negative before the stationary point and positive after it For a point of inflexion, the gradient has the same sign either side of the stationary point. See the Notes and Examples.

2. When testing the gradient either side of a stationary point, make sure the points you test are close enough to the stationary point you are investigating Otherwise, if there are two stationary points very close together, you may come to the wrong conclusion when identifying the stationary point. Example gradient negative here

gradient also negative here s

a

b

s is a minimum point between a and b, but the gradient is negative at a and negative at b. The gradient test would therefore suggest that s is a point of inflexion. This error is caused by there being two stationary points close together, both of which are between a and b.

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MEI C2 Differentiation Section 2 Crucial points 3. Remember that you can’t use the second derivative test when both the first and second derivatives are zero Example Find the nature of the turning point (0,0) on the curve y = x4.



Wrong

dy dy = 4x3. When x = 0, = 0. dx dx

d2 y d2 y 2 = 12x . When x = 0, = 0. dx 2 dx 2 So (0,0) is a point of inflection.



Right

x dy dx

We can’t tell the nature of the turning point in this case. We need to use the table method: 0 ve

0 0

0+ +ve

So (0, 0) is a local minimum point. 4. In a maximum or minimum problem, make sure that you answer the question For example, in a problem involving maximising the volume of a shape, make sure that you calculate the actual maximum volume obtained from the stationary value if you are asked for this.

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MEI Core 2 Differentiation Section 2: Applications of differentiation Exercise 1. A curve has equation y = x3 + 6x2 + 9x. (i)

Differentiate the function to obtain

dy . dx

(ii) Find the x co-ordinates of the points where

dy = 0 and hence the co-ordinates dx

of the turning points on the curve. dy (iii) By considering the sign of on either side of the turning points, determine dx whether the turning points are maximum or minimum points. (iv) Sketch the curve showing the turning points and points of intersection with the axes clearly.

2. The equation of a curve is given by y = 2x + x2 – 4x3. (i) Find the co-ordinates of the turning points on the curve, and distinguish between them by considering the gradient on either side of the turning points. (ii) Sketch the curve marking the turning points and points of intersection with the axes clearly. 3. A curve has equation y = x3 – 3x2 + 6. Find the co-ordinates of any turning points and determine their nature showing clearly how your decisions were made. Sketch the curve. 4. The equation of a curve is y = (x +1)(x – 3)3. (i) Write the equation of the curve in the form y = ax4 + bx3 + cx2 + dx + e. dy (ii) Find the co-ordinates of the points where = 0. dx (iii) Classify the stationary points. (iv) Sketch the curve. 5. Find the stationary points on the curve y = x4 – 2x3 and distinguish between them, showing all of the relevant working clearly. Sketch the curve. 6. The curve y = x3 + px2 + q has a minimum point at (4, -11). Find the co-ordinates of the maximum point on the curve. 7. The curve y = x3 + ax2 + bx + c passes through the point (1, 1). (i) Construct an equation connecting a, b and c. The curve also has turning points when x = -1 and when x = 3. (ii) Construct two further equations connecting a, b and c. (iii) Solve the three equations simultaneously to obtain values for a, b and c.

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MEI C2 Differentiation Section 2 Exercise 8. A variable rectangle has a constant perimeter of 20 cm. Find the lengths of the sides when the area is a maximum. 9. A square of side x cm is cut from the corners of a piece of card 15 cm by 24 cm. The card is then folded to form an open box. (i) Show that the volume of the box is (4x3 – 78x2 + 360x) cm3. (ii) Find a value for x that will make the volume a maximum. 10. A cylinder is cut from a solid sphere of radius 3cm. The height of the cylinder is 2h. (i) Find the radius of the cylinder in terms of h. (ii) Show that the volume of the cylinder is V = 2πh(9 – h2). (iii) Find the maximum volume of the cylinder as h varies.

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MEI Core 2 Differentiation Section 2: Applications of differentiation Solutions to Exercise 1. (i)

(ii)

y = x 3 + 6 x 2 + 9x dy = 3 x 2 + 12 x + 9 dx dy

=0 dx 3 x 2 + 12 x + 9 = 0

x 2 + 4x + 3 = 0 ( x + 1)( x + 3) = 0 x = −1 or x = −3 When x = -1, y = ( −1)3 + 6( −1)2 + 9 × −1 = −1 + 6 − 9 = −4 When x = -3, y = ( −3)3 + 6( −3)2 + 9 × −3 = −27 + 54 − 27 = 0 The turning points are (-1, -4) and (-3, 0) (iii) X dy

x< -3 +ve

x = -3 0

-3 < x < -1 -ve

x = -1 0

x > -1 +ve

dx The point (-3, 0) is a maximum point. The point(-1, -4) is a minimum point. (iv) y = x 3 + 6 x 2 + 9 x

= x( x 2 + 6 x + 9) = x( x + 3)2 The graph cuts the x-axis at x = 0 and x = -3 (repeated). The graph cuts the y-axis at y = 0.

-3

(-1, -4)

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MEI C2 Differentiation Section 2 Exercise solutions 2. (i) y = 2 x + x 2 − 4 x 3

dy dx

= 2 + 2 x − 12 x 2 dy

=0 dx 2 + 2 x − 12 x 2 = 0

At turning points,

1 + x − 6x 2 = 0 6x 2 − x − 1 = 0 (3 x + 1)(2 x − 1) = 0

x = − 31 or x = 21 2 3 4 = When x = − 31 , y = 2 ( − 31 ) + ( − 31 ) − 4 ( − 31 ) = − 23 + 19 + 27 When x = 21 , y = 2 ( 21 ) + ( 21 ) − 4 ( 21 ) = 1 + 41 − 21 = 2

3

11 The turning points are ( − 31 , − 27 ) and

x< − 31 -ve

x dy

x = − 31 0

−18 + 3 + 4 27

11 = − 27

3 4

( 21 , 43 ) .

− 31 < x < +ve

1 2

x= 0

1 2

x > 21 -ve

dx

( − 31 , − 2711 )

is a minimum point.

( , ) is a maximum point. 1 2

3 4

(ii) y = 2 x + x 2 − 4 x 3 = x(2 + x − 4 x 2 ) = − x(4 x 2 − x − 2) The curve cuts the x-axis at x = 0 and at the points satisfying 4x 2 − x − 2 = 0 . For this quadratic equation, a = 4,b = −1, c = −2 1 ± 1 − 4 × 4 × −2 1 ± 33 Using the quadratic formula, x = = 8 8

( 21 , 43 ) 1 8

(1 −

33 )

(−

1 3

,−

11 27

)

1 8

(1 +

33 )

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MEI C2 Differentiation Section 2 Exercise solutions 3. y = x 3 − 3 x 2 + 6 dy dx

= 3x 2 − 6x

At turning points,

dy

=0 dx 3x 2 − 6x = 0 3 x( x − 2) = 0 x = 0 or x = 2

When x = 0, y = 6 When x = 2, y = 2 3 − 3 × 2 2 + 6 = 8 − 12 + 6 = 2 The turning points are (0, 6) and (2, 2). d2 y dx 2

= 6x − 6

When x = 0, When x = 2,

d2 y dx 2 d2 y dx 2

= 0 − 6 < 0 , so (0, 6) is a maximum point. = 12 − 6 > 0 , so (2, 2) is a minimum point.

6 (2, 2)

4. (i)

y = ( x + 1)( x − 3)2 = ( x + 1)( x 2 − 6 x + 9) = x 3 − 6 x 2 + 9x + x 2 − 6 x + 9 = x 3 − 5 x 2 + 3x + 9

(ii) At turning points,

dy

=0 dx 3 x 2 − 10 x + 3 = 0 (3 x − 1)( x − 3) = 0

x=

or x = 3 When x = , y = ( + 1)( − 3)2 = 43 ×( − 83 )2 = 1 3

1 3

1 3

1 3

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MEI C2 Differentiation Section 2 Exercise solutions When x = 3, y = 0 The turning points are (iii) When x = 31 , When x = 3,

d2 y 2

dx d2 y

( 31 , 256 27 )

and (3, 0).

= 6 × 31 − 10 = 2 − 10 < 0 , so

( 31 , 256 27 )

is a maximum.

= 18 − 10 > 0 , so (3, 0) is a minimum.

dx 2

(iv) When y = 0, x = -1 or x = 3 When x = 0, y = 9 9

( 31 , 256 27 )

-1 3

5. y = x 4 − 2 x 3

dy

= 4x 3 − 6x 2

dx At stationary points, 4 x 3 − 6 x 2 = 0

x 2(2 x − 3) = 0 When x = 0, y = 0

x = 0 or x =

When x = 23 , y = ( 23 ) − 2 ( 23 ) = 4

x dy

3

x< 0 -ve

3 2

27 − 27 4 = − 16

81 16

x=0 0

0< x < -ve

3 2

x= 0

3 2

x > 23 +ve

dx So (0, 0) is a point of inflection, and

( 23 , − 2716 )

is a minimum point.

y

x

2

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MEI C2 Differentiation Section 2 Exercise solutions 6. y = x 3 + px 2 + q dy dx

= 3 x 2 + 2 px

At turning points,

dy

=0 dx 3 x 2 + 2 px = 0

x(3 x + 2 p ) = 0 x = 0 or x = −

2p 3

Since there is a minimum point at x = 4, − The curve is therefore y = x 3 − 6 x 2 + q .

2p = 4 ⇒ p = −6 3

The point (4, -11) lies on the curve, so −11 = 43 − 6 × 42 + q −11 = 64 − 96 + q ( 21 , 43 ) q = 21 3 The equation of the curve is y = x − 6 x 2 + 21 . The other turning point is at x = 0, so the maximum point is (0, 21).

7. (i)

y = x 3 + ax 2 + bx + c The graph passes through the point (1, 1) so 1 = 1 + a + b + c

a +b +c = 0 (ii)

dy

= 3 x 2 + 2ax + b

dx Turning points are when 3 x 2 + 2ax + b = 0 There is a turning point when x = -1, so 3( −1)2 + 2a × −1 + b = 0 3 − 2a + b = 0 2a − b = 3 There is a turning point when x = 3, so 3 × 3 2 + 2a × 3 + b = 0

27 + 6a + b = 0 6a + b = −27 (iii) a + b + c = 0 (1) 2a − b = 3 (2) 6a + b = −27 (3) Adding (2) and (3): Substituting into (2) gives: Substituting into (1) gives: a = −3, b = −9, c = 12

8a = −24 ⇒ a = −3 b = 2 a − 3 = −6 − 3 = − 9 c = −a − b = 9 + 3 = 12

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MEI C2 Differentiation Section 2 Exercise solutions 8. Let the length of the sides be x and y. Considering the perimeter: 2( x + y ) = 20 ⇒ x + y = 10 Let the area be A: A = xy

= x(10 − x ) = 10 x − x 2

dA = 10 − 2 x dx At turning point, 10 − 2 x = 0 2 x = 10 x =5 When x = 5, y = 10 – 5 = 5. d2 A = −2 so turning point is a maximum. dx 2 The area is a maximum when the lengths of the sides are 5 cm (i.e. the rectangle is a square). 9. (i) 24

x

x

24 - 2x

x

15 - 2x

15

x

Height of box is x cm Length of box is (24 – 2x) cm Width of box is (15 – 2x) cm Volume V = x(15 − 2 x )(24 − 2 x ) = x(360 − 78 x + 4 x 2 ) = 4 x 3 − 78 x 2 + 360 x

dV = 12 x 2 − 156 x + 360 dx At turning points, 12 x 2 − 156 x + 360 = 0

(ii)

x 2 − 13 x + 30 = 0 ( x − 3)( x − 10) = 0 x = 3 or x = 10

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MEI C2 Differentiation Section 2 Exercise solutions x = 10 is not possible since this would mean that the width would be negative. d2V = 24 x − 156 dx 2 d2V When x = 3, = 72 − 156 < 0 , so x = 1 is a maximum point. dx 2 The volume of the box is maximised when x = 3. (iii) Volume when x = 3 is V = 3 × 9 × 18 = 486 cm³

10. (i) 3

h r

r 2 + h 2 = 32 r =

9 −h2

(ii) Volume V = π r 2h = π ( 9 − h 2 ) × 2h = 2π h( 9 − h 2 )

(iii) V = 18π h − 2π h 3 dV = 18π − 6π h 2 dh At turning points, 18π − 6π h 2 = 0 3 −h2 = 0

h= 3 2

dV = −12π h so h = 3 is a maximum point. dh 2 Maximum volume V = 2π 3( 9 − 3) = 12π 3 cm³.

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MEI Core 2 Differentiation Chapter Assessment 1. Differentiate with respect to x: (i) x 5 (iii) 3

(ii) (iv)

2x 6 x 4  2 x  3x3

[2] [3]

2. Differentiate with respect to x: (i) y  2 x(3x 2  5) (ii)

y

[3]

x3  2 x 6 x2

[3]

3. Given that x = (3u + 2)(u2  3), find

d2 x in terms of u. du 2

[3]

4. A curve has equation y = x² – 3x + 1. (i) Find the equation of the tangent to the curve at the point where x = 1. (ii) Find the equation of the normal to the curve at the point where x = 3.

[3] [3]

5. A curve has equation y  2 x3  3x2  8x  9 . (i) Find the equation of the tangent to the curve at the point P (2, -3). [3] (ii) Find the coordinates of the point Q at which the tangent is parallel to the tangent at P. [3] 6. A curve has equation y = 2x3  6x. (i) Show that the curve crosses the x-axis at the origin and the points ( 3 , 0) and (  3 , 0) [2] dy (ii) Find . Hence find the stationary points on the curve. [3] dx d2 y (iii) Find , and use this to determine the nature of the stationary points. [3] dx 2 (iv) Sketch the curve. [2] 7. Two real numbers x and y are such that 2x + y = 100. Find the maximum value of the product of the two numbers. [5] 8. A cuboid has a square base of length x cm and height y cm. (i) Express the volume V of the cuboid in terms of x and y. (ii) Show that the surface area of the cuboid is 2x2 + 4xy.

[1] [2]

The surface area of the cuboid is 24 cm2. (iii) Show that V = 6x  0.5x3 [2] (iv) Show that the maximum volume of the box occurs when x = 2, and find this maximum volume. [4] Total 50 marks

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MEI Core 2 Differentiation Section 2: Applications of differentiation Multiple Choice Test 1. The point (a, b) is a local maximum if

dy d2 y = 0 and dx 2 dx dy d2 y (b) When x = a, = 0 and 2 dx dx 2 dy d y (c) When x = a, 2 = 0 and dx dx 2 dy d y (d) When x = a, = 0 and 2 dx dx (e) I don’t know (a) When x = a,

is negative is positive is positive is negative

d2 y 2. Given that y = 3x(1  x ), 2 = dx 3

(a) 36x2 (c) 0 (e) I don’t know

(b) 3  12x3 (d) 18x2

3. The turning point(s) of the curve y = 3x  x3 are (a) (1, 2) max (c) (2, 1) max and (2, 1) min (e) I don’t know

4. If, when x = a,

(b) (1, 2) min and (1, 2) max (d) (1, 2) max and (1, 2) min

dy d2 y = 0 and 2 = 0: dx dx

(a) x = a is a local minimum (b) x = a is a local maximum (c) x = a is a stationary point but could be a local maximum, local minimum or a stationary point of inflection (d) x = a is a stationary point of inflection (e) I don’t know

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MEI C2 Differentiation Section 2 MC test 5. The displacement s metres of a particle after t seconds is given by s = t3  6t. d2s ds The velocity v and the acceleration a can be found using v  and a  2 . dt dt 1 2 After 2 seconds, the velocity v m s and acceleration a m s are given by: (a) v = 30 and a = 12 (c) v = 12 and a = 6 (e) I don’t know

(b) v = 6 and a = 0 (d) v = 6 and a = 12

6. Given that x + y = 60, the maximum value of x2y is (a) 40 (c) 16000 (e) I don’t know

(b) 32000 (d) 20

Questions 7 and 8 are about a straight wall AB and a fence of length 5 m which form a rectangular enclosure. The width of the enclosure is x m. A

B xm

7. The area of the enclosure, in m², is given by (a) x(5  x) (c) x(5  2 x) (e) I don’t know

(b) x(2 x  5) (d) x( x  5)

8. The maximum value of the area of the enclosure is (a) 3.125 m² (c) 6.25 m² (e) I don’t know

(b) 1.25 m² (d) 4.6875 m²

9. The gradient of the chord joining (x, y) to (x + x, y + y) on the curve y = 5x2 is:

5( x   ) 2  5 x 2 x dy (c) dx (e) I don’t know (a)

(b)

5( x  x) 2  5 x 2 y

5( x  x) 2  5 x 2 (d) x

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MEI C2 Differentiation Section 2 MC test 10. The gradient of the chord joining (x, y) to (x + x, y + y) on the curve y = x2 + 3x is: (a) 2x (c) 2x + 3 + x (e) I don’t know

dy dx (d) 2x + 3 + y (b)

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