Revision Library

revisionlibrary.co.uk

GCSE · Calculus

C2 Applications of Differentiation: Gradients, Tangents and Turning Points

AQA 8365 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A curve has equation y = x³ + 2x² - 5x + 7.

(a)Find dy/dx.(2)

(b)Find the gradient of the curve at the point where x = 2.(2)

(Total for Question 1 is 4 marks)

A curve has equation y = 4x² - 3x + 1.

(a)Find dy/dx.(2)

(b)Find the equation of the tangent to the curve at the point where x = 1. Give your answer in the form y = mx + c.(3)

(Total for Question 2 is 5 marks)

A curve has equation y = 3x² + 4/x, for x not equal to 0.

(a)Find dy/dx.(2)

(b)Find the gradient of the curve at the point where x = 2.(2)

(Total for Question 3 is 4 marks)

A curve has equation y = x² - 4x + 5. Find the equation of the normal to the curve at the point where x = 3. Give your answer in the form x + 2y = c.

(Total for Question 4 is 4 marks)

A curve has equation y = x² - 6x + 10.

(a)Find dy/dx.(1)

(b)Find the coordinates of the turning point of the curve, and state whether it is a maximum or a minimum point.(3)

(Total for Question 5 is 4 marks)

A curve C has equation y = 2x³ - 3x² - 12x + 5. Show that x = 2 is a stationary point of C.

(Total for Question 6 is 3 marks)

A ball is thrown vertically upwards. Its height, h metres, above the ground after t seconds is modelled by h = 20t - 5t², for t ≥ 0.

(a)Find dh/dt. This represents the vertical velocity of the ball.(1)

(b)Find the time at which the ball reaches its maximum height, and find this maximum height.(4)

(c)Find d2h/dt2 (the acceleration of the ball), and explain what this value represents in this context.(2)

(Total for Question 7 is 7 marks)

The temperature, T degrees Celsius, of a chemical reaction t minutes after it begins is modelled by T = -2t³ + 21t² - 60t + 40, for 0 ≤ t ≤ 8.

(a)Find dT/dt.(1)

(b)Find the values of t, where 0 ≤ t ≤ 8, at which the temperature is momentarily not changing.(3)

(c)Determine, using the second derivative, whether the temperature is a maximum or a minimum at each of these times.(3)

(Total for Question 8 is 7 marks)

A curve has equation y = x³ - 6x² - 15x + 3.

(a)Find dy/dx.(1)

(b)Find the x-coordinates of the stationary points of the curve.(3)

(c)Find the corresponding y-coordinates of each stationary point.(2)

(d)Use the second derivative to determine the nature of each stationary point.(3)

(Total for Question 9 is 9 marks)

A gardener has 60 metres of edging to enclose a rectangular flower bed on all four sides. Let x metres be the width of the flower bed and A m² be the area enclosed.

(a)Show that A = 30x - x^2.(2)

(b)Find dA/dx.(1)

(c)Find the value of x that gives the maximum area, and calculate this maximum area.(4)

(Total for Question 10 is 7 marks)

Find the equation of the tangent to the curve y = x² - 5x + 3 that is parallel to the line y = 3x - 7.

(Total for Question 11 is 5 marks)

The profit, P thousand pounds, made by a company from selling x thousand items of a product is modelled by P = -2x² + 40x - 50, for x > 0.

(a)Find dP/dx.(1)

(b)Find the number of items (in thousands) that should be sold to maximise profit, and find this maximum profit.(4)

(c)Use the second derivative to justify that this value of x gives a maximum profit.(2)

(Total for Question 12 is 7 marks)

A curve has equation y = x² - 2x - 3.

(a)Find the equation of the normal to the curve at the point where x = 0.(4)

(b)Find the area of the triangle enclosed by this normal and the two coordinate axes.(3)

(Total for Question 13 is 7 marks)

A curve has equation y = x³ - 9x² + 24x - 5.

(a)Find dy/dx.(1)

(b)Find the coordinates of the stationary points of the curve.(4)

(c)Use the second derivative to determine the nature of each stationary point.(3)

(d)Find the range of values of x for which the curve is a decreasing function.(2)

(Total for Question 14 is 10 marks)

The curve C has equation y = x² - 4x + 7. The point P(1, 4) lies on C.

(a)Confirm that P lies on C, and find the equation of the tangent to C at P.(4)

(b)The tangent meets the y-axis at the point Q. Find the coordinates of Q.(1)

(c)Find the area of triangle OPQ, where O is the origin.(3)

(Total for Question 15 is 8 marks)

A curve has equation y = (x² + 6)/x, for x not equal to 0.

(a)Show that y can be written as x + 6/x.(1)

(b)Find dy/dx.(2)

(c)Find the coordinates of the stationary points of the curve.(4)

(d)Use the second derivative to determine the nature of each stationary point.(3)

(Total for Question 16 is 10 marks)

A rectangular sheet of card measures 24 cm by 15 cm. A square of side x cm is cut from each corner, and the sides are folded up to form an open-topped box.

(a)Show that the volume, V cm³, of the box is given by V = 4x³ - 78x² + 360x.(3)

(b)Find dV/dx.(2)

(c)Given that 0 < x < 7.5, find the value of x that maximises the volume of the box, and find this maximum volume.(5)

(Total for Question 17 is 10 marks)

A curve has equation y = x³ - 3kx + 2, where k is a non-zero constant.

(a)Find dy/dx in terms of k.(2)

(b)Show that the curve has two distinct stationary points only when k > 0, and find their x-coordinates in terms of k.(4)

(Total for Question 18 is 6 marks)

Curve C1 has equation y = x² - 2x + 5. Curve C2 has equation y = -x² + 6x - 3.

(a)Find the gradient function of each curve.(2)

(b)Find the value of x for which the two curves have the same gradient.(2)

(c)Show that the two curves also meet at this value of x, and state what this means geometrically.(3)

(Total for Question 19 is 7 marks)

A curve has equation y = 2x³ - 15x² + 24x + 6.

(a)Find dy/dx.(1)

(b)Find the x-coordinates of the stationary points of the curve.(3)

(c)Find the corresponding y-coordinates of each stationary point.(2)

(d)Use the second derivative to determine the nature of each stationary point.(3)

(e)Find the range of values of x for which the curve is an increasing function.(2)

(Total for Question 20 is 11 marks)

Mark scheme · C2 Applications of Differentiation: Gradients, Tangents and Turning Points

Question 1

(a) M1 differentiates at least two terms correctly (e.g. 3x² and 4x seen)
(a) A1 dy/dx = 3x² + 4x - 5 cao
(a) Answer: dy/dx = 3x² + 4x - 5
(b) M1 substitutes x = 2 into their dy/dx, ft from part (a)
(b) A1 gradient = 15 cao
(b) Answer: 15

Question 2

(a) M1 differentiates at least one term correctly
(a) A1 dy/dx = 8x - 3 cao
(a) Answer: dy/dx = 8x - 3
(b) M1 finds the y-coordinate at x = 1, y = 2
(b) M1 finds the gradient at x = 1 using their dy/dx, gradient = 5, ft from part (a)
(b) A1 y = 5x - 3 oe
(b) Answer: y = 5x - 3

Question 3

(a) M1 writes 4/x as 4x^-1 and differentiates at least one term correctly
(a) A1 dy/dx = 6x - 4/x² oe cao
(a) Answer: dy/dx = 6x - 4/x²
(b) M1 substitutes x = 2 into their dy/dx, ft from part (a)
(b) A1 gradient = 11 cao
(b) Answer: 11

Question 4

M1 finds the y-coordinate at x = 3, y = 2
M1 finds the gradient of the tangent at x = 3 using dy/dx = 2x - 4, gradient = 2
M1 uses normal gradient = -1/(tangent gradient) = -1/2, dependent on previous M1 (dM1)
A1 x + 2y = 7 oe (e.g. y = -0.5x + 3.5)
Answer: x + 2y = 7

Question 5

(a) B1 dy/dx = 2x - 6 cao
(a) Answer: dy/dx = 2x - 6
(b) M1 sets their dy/dx = 0 and solves for x, ft from part (a)
(b) A1 turning point (3, 1)
(b) A1 minimum, since the coefficient of x² is positive (the curve is u-shaped) oe
(b) Answer: (3, 1), a minimum point

Question 6

M1 differentiates correctly to obtain dy/dx = 6x² - 6x - 12
M1 substitutes x = 2 into dy/dx
A1 dy/dx = 0 at x = 2, so x = 2 is a stationary point, fully shown cso
Answer: dy/dx = 0 at x = 2 (shown)

Question 7

(a) B1 dh/dt = 20 - 10t cao
(a) Answer: dh/dt = 20 - 10t
(b) M1 sets their dh/dt = 0, ft from part (a)
(b) A1 t = 2 seconds
(b) M1 substitutes t = 2 into h = 20t - 5t², dependent on previous M1 (dM1)
(b) A1 maximum height = 20 metres
(b) Answer: t = 2 seconds, maximum height = 20 m
(c) B1 d2h/dt2 = -10
(c) B1 correct interpretation: this is the (constant) acceleration due to gravity, acting downwards, causing the ball to decelerate on the way up oe
(c) Answer: -10 m/s², the constant deceleration due to gravity

Question 8

(a) B1 dT/dt = -6t² + 42t - 60 cao
(a) Answer: dT/dt = -6t² + 42t - 60
(b) M1 sets their dT/dt = 0 and simplifies, e.g. dividing by -6 to give t² - 7t + 10 = 0, ft from part (a)
(b) M1 factorises or solves the quadratic, e.g. (t-2)(t-5) = 0
(b) A1 t = 2 and t = 5 (both values) cao
(b) Answer: t = 2 and t = 5
(c) M1 finds the second derivative d2T/dt2 = -12t + 42
(c) A1 evaluates at both t = 2 (= 18) and t = 5 (= -18), ft from part (b)
(c) A1 t = 2 gives a minimum temperature (d2T/dt2 = 18 > 0); t = 5 gives a maximum temperature (d2T/dt2 = -18 < 0)
(c) Answer: Minimum at t = 2 (T = -12 C); maximum at t = 5 (T = 15 C)

Question 9

(a) B1 dy/dx = 3x² - 12x - 15 cao
(a) Answer: dy/dx = 3x² - 12x - 15
(b) M1 sets their dy/dx = 0 and simplifies, e.g. dividing by 3 to give x² - 4x - 5 = 0, ft from part (a)
(b) M1 factorises or solves the quadratic, e.g. (x-5)(x+1) = 0
(b) A1 x = 5 and x = -1 (both values) cao
(b) Answer: x = 5 and x = -1
(c) M1 substitutes both x-values into y = x³ - 6x² - 15x + 3, ft from part (b)
(c) A1 (5, -97) and (-1, 11) (both required) cao
(c) Answer: (5, -97) and (-1, 11)
(d) M1 finds the second derivative d2y/dx2 = 6x - 12
(d) A1 evaluates at both x = 5 (= 18) and x = -1 (= -18), ft from part (b)
(d) A1 (5, -97) is a minimum (d2y/dx2 = 18 > 0); (-1, 11) is a maximum (d2y/dx2 = -18 < 0)
(d) Answer: Minimum at (5, -97); maximum at (-1, 11)

Question 10

(a) M1 uses the perimeter 2x + 2y = 60 to express the length as y = 30 - x, and multiplies by x
(a) A1 A = 30x - x² fully shown cso
(a) Answer: A = 30x - x² (shown)
(b) B1 dA/dx = 30 - 2x cao, ft from part (a)
(b) Answer: dA/dx = 30 - 2x
(c) M1 sets their dA/dx = 0 and solves for x, ft from part (b)
(c) A1 x = 15
(c) M1 substitutes x = 15 back into A = 30x - x², dependent on previous M1 (dM1)
(c) A1 maximum area = 225 m²
(c) Answer: x = 15 m, maximum area = 225 m²

Question 11

M1 identifies that the gradient of the line y = 3x - 7 is 3
M1 differentiates the curve and sets dy/dx = 2x - 5 = 3, dependent on previous M1 (dM1)
A1 x = 4
A1 y = -1, giving the point (4, -1)
A1 y = 3x - 13 oe
Answer: y = 3x - 13

Question 12

(a) B1 dP/dx = -4x + 40 cao
(a) Answer: dP/dx = -4x + 40
(b) M1 sets their dP/dx = 0 and solves for x, ft from part (a)
(b) A1 x = 10
(b) M1 substitutes x = 10 back into P = -2x² + 40x - 50, dependent on previous M1 (dM1)
(b) A1 maximum profit = 150 (i.e. GBP 150,000)
(b) Answer: x = 10 (10,000 items), maximum profit = GBP 150,000
(c) B1 d2P/dx2 = -4
(c) B1 since d2P/dx2 < 0, this confirms a maximum
(c) Answer: d2P/dx2 = -4 < 0, so x = 10 gives a maximum

Question 13

(a) M1 finds the y-coordinate at x = 0, y = -3
(a) M1 finds the gradient of the tangent at x = 0 using dy/dx = 2x - 2, gradient = -2
(a) M1 uses normal gradient = -1/(tangent gradient) = 1/2, dependent on previous M1 (dM1)
(a) A1 y = 0.5x - 3 oe
(a) Answer: y = 0.5x - 3
(b) M1 finds the x-intercept of the normal, ft from part (a)
(b) M1 identifies the two perpendicular distances (base and height) from the axis intercepts, dependent on previous M1 (dM1)
(b) A1 area = 9 square units
(b) Answer: 9 square units

Question 14

(a) B1 dy/dx = 3x² - 18x + 24 cao
(a) Answer: dy/dx = 3x² - 18x + 24
(b) M1 sets their dy/dx = 0 and simplifies, e.g. dividing by 3 to give x² - 6x + 8 = 0, ft from part (a)
(b) M1 factorises or solves the quadratic, e.g. (x-2)(x-4) = 0
(b) A1 x = 2 and x = 4 (both values) cao
(b) A1 (2, 15) and (4, 11) (both coordinate pairs) cao
(b) Answer: (2, 15) and (4, 11)
(c) M1 finds the second derivative d2y/dx2 = 6x - 18
(c) A1 evaluates at both x = 2 (= -6) and x = 4 (= 6), ft from part (b)
(c) A1 (2, 15) is a maximum (d2y/dx2 = -6 < 0); (4, 11) is a minimum (d2y/dx2 = 6 > 0)
(c) Answer: Maximum at (2, 15); minimum at (4, 11)
(d) M1 tests the sign of dy/dx = 3(x-2)(x-4) between the stationary values
(d) A1 2 < x < 4 oe
(d) Answer: 2 < x < 4

Question 15

(a) B1 confirms y(1) = 1-4+7 = 4, so P lies on C
(a) M1 differentiates to obtain dy/dx = 2x - 4, and finds the gradient at x = 1, gradient = -2
(a) M1 forms the tangent equation y - 4 = -2(x - 1), dependent on previous M1 (dM1)
(a) A1 y = -2x + 6 oe
(a) Answer: y = -2x + 6
(b) B1 Q = (0, 6), ft from part (a)
(b) Answer: Q = (0, 6)
(c) M1 identifies OQ = 6 as a suitable base (on the y-axis), ft from part (b)
(c) M1 identifies the height as the horizontal distance from P to the y-axis, height = 1, dependent on previous M1 (dM1)
(c) A1 area = 3 square units
(c) Answer: 3 square units

Question 16

(a) B1 y = x²/x + 6/x = x + 6/x, fully shown cso
(a) Answer: y = x + 6/x (shown)
(b) M1 writes 6/x as 6x^-1 and differentiates
(b) A1 dy/dx = 1 - 6/x² oe cao
(b) Answer: dy/dx = 1 - 6/x²
(c) M1 sets their dy/dx = 0, giving x² = 6, ft from part (b)
(c) M1 takes both square roots, x = √6 or x = -√6
(c) A1 x = √6 and x = -√6 (awrt 2.45 and -2.45) cao
(c) A1 (√6, 2sqrt(6)) and (-√6, -2sqrt(6)) (awrt (2.45, 4.90) and (-2.45, -4.90))
(c) Answer: (√6, 2sqrt(6)) and (-√6, -2sqrt(6)), i.e. approximately (2.45, 4.90) and (-2.45, -4.90)
(d) M1 finds the second derivative d2y/dx2 = 12/x³, ft from part (b)
(d) A1 evaluates the sign at x = √6 (positive) and at x = -√6 (negative)
(d) A1 (√6, 2sqrt(6)) is a minimum; (-√6, -2sqrt(6)) is a maximum
(d) Answer: Minimum at (√6, 2sqrt(6)); maximum at (-√6, -2sqrt(6))

Question 17

(a) M1 writes the base dimensions as (24 - 2x) and (15 - 2x), with height x
(a) M1 expands V = x(24-2x)(15-2x) to obtain 360x - 78x² + 4x³ (or equivalent unsimplified form)
(a) A1 V = 4x³ - 78x² + 360x, fully shown cso
(a) Answer: V = 4x³ - 78x² + 360x (shown)
(b) M1 differentiates at least two terms correctly, ft from part (a)
(b) A1 dV/dx = 12x² - 156x + 360 cao
(b) Answer: dV/dx = 12x² - 156x + 360
(c) M1 sets their dV/dx = 0 and simplifies, e.g. dividing by 12 to give x² - 13x + 30 = 0, ft from part (b)
(c) M1 factorises or solves the quadratic, e.g. (x-3)(x-10) = 0
(c) A1 rejects x = 10 as it lies outside 0 < x < 7.5, so x = 3
(c) M1 substitutes x = 3 into V = 4x³ - 78x² + 360x, dependent on previous M1 (dM1)
(c) A1 maximum volume = 486 cm³
(c) Answer: x = 3 cm, maximum volume = 486 cm³

Question 18

(a) M1 differentiates the term -3kx correctly
(a) A1 dy/dx = 3x² - 3k cao
(a) Answer: dy/dx = 3x² - 3k
(b) M1 sets their dy/dx = 0 to obtain x² = k, ft from part (a)
(b) M1 considers the cases k > 0, k = 0 and k < 0 for real solutions of x² = k
(b) A1 correct reasoning: k < 0 gives no real solutions; k = 0 gives one repeated solution x = 0; k > 0 gives two distinct real solutions, so two distinct stationary points require k > 0, cso
(b) A1 x = √k and x = -√k
(b) Answer: Two distinct stationary points exist only when k > 0, at x = √k and x = -√k

Question 19

(a) B1 dy1/dx = 2x - 2
(a) B1 dy2/dx = -2x + 6
(a) Answer: dy1/dx = 2x - 2; dy2/dx = -2x + 6
(b) M1 sets 2x - 2 = -2x + 6, ft from part (a)
(b) A1 x = 2 cao
(b) Answer: x = 2
(c) M1 substitutes x = 2 into both y1 and y2, ft from part (b)
(c) A1 y1(2) = 5 and y2(2) = 5, so both curves pass through (2, 5), fully shown cso
(c) A1 correct geometric statement: since the curves meet at (2, 5) and have the same gradient there, they touch (are tangent to each other) at this point oe
(c) Answer: Both curves pass through (2, 5) with the same gradient, so the curves are tangent to each other at (2, 5)

Question 20

(a) B1 dy/dx = 6x² - 30x + 24 cao
(a) Answer: dy/dx = 6x² - 30x + 24
(b) M1 sets their dy/dx = 0 and simplifies, e.g. dividing by 6 to give x² - 5x + 4 = 0, ft from part (a)
(b) M1 factorises or solves the quadratic, e.g. (x-1)(x-4) = 0
(b) A1 x = 1 and x = 4 (both values) cao
(b) Answer: x = 1 and x = 4
(c) M1 substitutes both x-values into y = 2x³ - 15x² + 24x + 6, ft from part (b)
(c) A1 (1, 17) and (4, -10) (both required) cao
(c) Answer: (1, 17) and (4, -10)
(d) M1 finds the second derivative d2y/dx2 = 12x - 30
(d) A1 evaluates at both x = 1 (= -18) and x = 4 (= 18), ft from part (b)
(d) A1 (1, 17) is a maximum (d2y/dx2 = -18 < 0); (4, -10) is a minimum (d2y/dx2 = 18 > 0)
(d) Answer: Maximum at (1, 17); minimum at (4, -10)
(e) M1 tests the sign of dy/dx = 6(x-1)(x-4) outside the stationary values
(e) A1 x < 1 or x > 4 oe
(e) Answer: x < 1 or x > 4

Applications of Differentiation: Gradients, Tangents and Turning Points - Worksheets, Questions and Revision

C2 Applications of Differentiation: Gradients, Tangents and Turning Points

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20