Pure: Differentiation Depth - Worksheets, Questions and Revision

12 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 4)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

A-Level · Pure Mathematics

P14 Pure: Differentiation Depth

EDEXCEL 9MA0 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A curve has equation y = (3x² - 5)⁴.

(a)Find dy/dx, giving your answer in fully simplified factorised form.(3)

(b)Hence find the exact value of dy/dx when x = 2.(2)

(Total for Question 1 is 5 marks)

A curve has equation y = x² * e^3x.

(a)Show that dy/dx can be written as x * e^3x * (3x + 2).(4)

(b)Hence find the exact value(s) of x for which the tangent to the curve is horizontal.(2)

(Total for Question 2 is 6 marks)

A curve has equation y = (2x - 1)/(x² + 4).

(a)Find dy/dx, giving your answer as a single simplified fraction.(3)

(b)Hence find the exact x-coordinates of the points where dy/dx = 0, giving your answers in the form (a ± √b)/c.(2)

(Total for Question 3 is 5 marks)

A curve has equation y = e^cos(2x).

(a)Find dy/dx.(2)

(b)Find the exact value of dy/dx when x = π/6, giving your answer in the form -√ke for an integer k.(3)

(Total for Question 4 is 5 marks)

A curve has equation y = x² * e^-x.

(a)Find dy/dx, giving your answer fully factorised.(3)

(b)Show that the curve has stationary points where x = 0 and x = 2, and find the corresponding y-coordinates.(3)

(c)Find d²y/dx² and hence determine the nature of each stationary point.(2)

(Total for Question 5 is 8 marks)

A curve C has equation x² + y² - 4x + 6y = 12.

(a)Show that dy/dx = (2 - x)/(y + 3).(3)

(b)Verify that the point (2, 2) lies on C, and find the gradient of C at this point.(2)

(Total for Question 6 is 5 marks)

A curve C is defined by the equation x²*y - y² = 14. The point P(3, 2) lies on C.

(a)Find dy/dx in terms of x and y.(4)

(b)Show that the gradient of C at P is -12/5.(2)

(c)Find an equation of the tangent to C at P, giving your answer in the form ax + by + c = 0 with integer coefficients.(3)

(d)Find an equation of the normal to C at P, giving your answer in the form ax + by + c = 0 with integer coefficients.(3)

(Total for Question 7 is 12 marks)

A curve has parametric equations x = t² - 1, y = 3t - t³, for t in R.

(a)Find dy/dx in terms of t.(3)

(b)Find the value of dy/dx when t = 2.(2)

(Total for Question 8 is 5 marks)

A curve C has parametric equations x = 2t + 1, y = t² - 4t, for t in R.

(a)Find dy/dx in terms of t.(3)

(b)Find the coordinates of the point on C where the tangent is parallel to the line y = 2x + 5.(4)

(c)Show that the cartesian equation of C can be written in the form y = ax² + bx + c, stating the values of a, b and c.(3)

(Total for Question 9 is 10 marks)

A spherical balloon is inflated so that its volume V cm³ increases at a constant rate of 200 cm³ per second. At time t seconds the radius of the balloon is r cm.

(a)Show that dr/dt = 50/(π*r²).(3)

(b)Find the rate of increase of the radius at the instant when r = 10 cm, giving your answer to 3 significant figures.(2)

(c)Find the rate of increase of the surface area of the balloon at this instant.(3)

(Total for Question 10 is 8 marks)

Water is poured into an inverted right circular conical tank at a constant rate of 30 cm³ per second. The tank has height 40 cm and the radius of its circular top is 20 cm, so the radius r cm and depth h cm of the water are always in the same ratio as the tank's dimensions. Let V cm³ be the volume of water in the tank when the depth is h cm.

(a)Show that V = (π/12)*h³.(3)

(b)Find dh/dt in terms of h.(3)

(c)Find the rate at which the depth is increasing at the instant when h = 10 cm, giving your answer to 3 significant figures.(3)

(d)Explain, with justification, whether the rate at which the depth is increasing gets faster or slower as h increases.(2)

(Total for Question 11 is 11 marks)

A uniform ladder of length 5 m rests with its foot on horizontal ground and its top against a vertical wall. The foot of the ladder is pulled directly away from the wall at a constant rate of 0.4 m per second. At time t seconds, let x m be the distance of the foot from the wall and let θ radians be the angle between the ladder and the ground, so that x = 5*cos(θ).

(a)Show that dx/dt = -5*sin(θ)*(dtheta/dt).(2)

(b)Hence find the exact rate of change of θ at the instant when θ = π/3, and interpret the sign of your answer.(5)

(Total for Question 12 is 7 marks)

Mark scheme · P14 Pure: Differentiation Depth

Question 1

(a) M1 attempt at the chain rule, differentiating to the form k(3x²-5)³
(a) A1 correct unsimplified derivative 4(3x²-5)³ * 6x identified
(a) A1 dy/dx = 24x(3x²-5)³ cao
(a) Answer: dy/dx = 24x(3x² - 5)³
(b) M1 substitute x=2 into their dy/dx expression (ft)
(b) A1 16464 cao
(b) Answer: 16464

Question 2

(a) M1 attempt at the product rule with u=x², v=e^3x
(a) A1 correct unsimplified derivative 2x*e^3x + 3x²*e^3x
(a) dM1 factorise out the common factor x*e^3x, dependent on the previous M1
(a) A1 dy/dx = x*e^3x*(3x+2) cso
(a) Answer: dy/dx = x * e^3x * (3x + 2)
(b) M1 set dy/dx=0 and use that e^3x is never zero to obtain x(3x+2)=0
(b) A1 x = 0 or x = -2/3 both stated oe
(b) Answer: x = 0 or x = -2/3

Question 3

(a) M1 attempt at the quotient rule in the form (v*u' - u*v')/v² or equivalent
(a) A1 correct unsimplified numerator 2(x²+4) - (2x-1)(2x)
(a) A1 dy/dx = (-2x²+2x+8)/(x²+4)² oe cao, e.g. -2(x²-x-4)/(x²+4)²
(a) Answer: dy/dx = (-2x² + 2x + 8)/(x²+4)², equivalently -2(x² - x - 4)/(x²+4)²
(b) M1 set the numerator -2x²+2x+8=0 (i.e. x²-x-4=0) and apply the quadratic formula
(b) A1 x = (1+√17)/2 or x = (1-√17)/2, both stated oe
(b) Answer: x = (1 + √17)/2 or x = (1 - √17)/2 (awrt 2.56 or -1.56)

Question 4

(a) M1 attempt at the chain rule to the form k*sin(2x)*e^cos(2x)
(a) A1 dy/dx = -2*sin(2x)*e^cos(2x) cao
(a) Answer: dy/dx = -2 * sin(2x) * e^cos(2x)
(b) M1 substitute x=π/6 using the exact values sin(π/3)=√3/2 and cos(π/3)=1/2
(b) A1 correct unsimplified exact value -√3 * e^1/2
(b) A1 dy/dx = -√3e cao
(b) Answer: dy/dx = -√3e, equivalently -√3 * e^1/2 (awrt -2.856)

Question 5

(a) M1 attempt at the product rule with u=x², v=e^-x
(a) A1 correct unsimplified derivative 2x*e^-x - x²*e^-x
(a) A1 dy/dx = x*e^-x*(2-x) cao
(a) Answer: dy/dx = x * e^-x * (2 - x)
(b) M1 set dy/dx=0 and use that e^-x is never zero to obtain x(2-x)=0
(b) A1 x=0 and x=2 both stated cso
(b) A1 y-coordinates y=0 (at x=0) and y=4/e², awrt 0.541 (at x=2), both cao
(b) Answer: Stationary points at (0, 0) and (2, 4/e²), where 4/e² is awrt 0.541
(c) M1 correct d²y/dx² = e^-x*(x²-4x+2), found by differentiating dy/dx again using the product rule
(c) A1 correct classification with reasoning: (0,0) is a minimum since d²y/dx²=2>0 at x=0; (2, 4/e²) is a maximum since d²y/dx²=-2/e²<0 at x=2, both cao
(c) Answer: (0, 0) is a minimum point; (2, 4/e²) is a maximum point

Question 6

(a) M1 differentiate implicitly term by term, with at least 3 of the 4 terms correct (2x, 2y*dy/dx, -4, 6*dy/dx)
(a) A1 fully correct implicit equation 2x + 2y*(dy/dx) - 4 + 6*(dy/dx) = 0
(a) A1 rearrange to dy/dx = (2-x)/(y+3) cso
(a) Answer: dy/dx = (2 - x)/(y + 3)
(b) M1 verify (2,2) satisfies the equation (4+4-8+12=12) and substitute x=2, y=2 into the implicit derivative
(b) A1 dy/dx = 0 cao
(b) Answer: Gradient at (2, 2) is 0 (a horizontal tangent)

Question 7

(a) M1 differentiate x²*y implicitly using the product rule to obtain 2xy + x²*(dy/dx)
(a) M1 differentiate y² implicitly to obtain 2y*(dy/dx)
(a) A1 correct full implicit equation 2xy + x²*(dy/dx) - 2y*(dy/dx) = 0
(a) A1 dy/dx = 2xy/(2y-x²) oe cao, e.g. -2xy/(x²-2y)
(a) Answer: dy/dx = 2xy/(2y - x²)
(b) M1 substitute x=3, y=2 into their dy/dx expression (ft)
(b) A1 gradient = -12/5 cso
(b) Answer: Gradient at P = -12/5
(c) M1 use y-2 = m(x-3) with m=-12/5 (ft their gradient from part (b))
(c) M1 multiply through by 5 and rearrange to clear fractions
(c) A1 12x + 5y - 46 = 0 cao (oe, any correct integer multiple)
(c) Answer: 12x + 5y - 46 = 0
(d) M1 normal gradient = 5/12 (negative reciprocal of their tangent gradient, ft)
(d) M1 use y-2 = (5/12)(x-3) and clear fractions
(d) A1 5x - 12y + 9 = 0 cao (oe)
(d) Answer: 5x - 12y + 9 = 0

Question 8

(a) M1 dx/dt = 2t
(a) M1 dy/dt = 3 - 3t²
(a) A1 dy/dx = (3-3t²)/(2t) oe cao, e.g. 3(1-t²)/(2t)
(a) Answer: dy/dx = (3 - 3t²)/(2t)
(b) M1 substitute t=2 into their dy/dx expression (ft)
(b) A1 dy/dx = -9/4 oe cao (accept -2.25)
(b) Answer: dy/dx = -9/4

Question 9

(a) M1 dx/dt = 2
(a) M1 dy/dt = 2t-4
(a) A1 dy/dx = t-2 oe cao
(a) Answer: dy/dx = t - 2
(b) M1 set dy/dx equal to the gradient of the given line (2)
(b) A1 t=4
(b) M1 substitute t=4 into both parametric equations
(b) A1 point (9, 0) cao
(b) Answer: (9, 0)
(c) M1 make t the subject of the x equation, t=(x-1)/2
(c) M1 substitute into y=t²-4t and expand
(c) A1 y = (1/4)x² - (5/2)x + 9/4 cso (equivalently 4y=x²-10x+9), with a=1/4, b=-5/2, c=9/4 stated
(c) Answer: y = (1/4)x² - (5/2)x + 9/4, so a = 1/4, b = -5/2, c = 9/4

Question 10

(a) M1 differentiate V=(4/3)*π*r³ to obtain dV/dr = 4*π*r²
(a) M1 use the chain rule dr/dt = (dV/dt)/(dV/dr) with dV/dt=200
(a) A1 dr/dt = 200/(4*π*r²) = 50/(π*r²) cso
(a) Answer: dr/dt = 50/(π*r²)
(b) M1 substitute r=10 into the formula from part (a)
(b) A1 dr/dt = awrt 0.159 cm/s cao
(b) Answer: dr/dt = 1/(2*π) cm/s, awrt 0.159 cm/s
(c) M1 differentiate S=4*π*r² to obtain dS/dr = 8*π*r
(c) M1 use the chain rule dS/dt = (dS/dr)*(dr/dt) with their dr/dt (ft)
(c) A1 dS/dt = 40 cm² per second cao
(c) Answer: dS/dt = 40 cm² per second

Question 11

(a) M1 use similar triangles to write r in terms of h, r = h/2
(a) M1 substitute into V=(1/3)*π*r²*h
(a) A1 V = (π/12)*h³ cso, with full simplification shown
(a) Answer: V = (π/12)*h³
(b) M1 differentiate V=(π/12)*h³ to obtain dV/dh = (π/4)*h²
(b) M1 use the chain rule dh/dt = (dV/dt)/(dV/dh) with dV/dt=30
(b) A1 dh/dt = 120/(π*h²) oe cao
(b) Answer: dh/dt = 120/(π*h²)
(c) M1 substitute h=10 into their dh/dt expression (ft)
(c) A1 exact value 6/(5*π) oe, e.g. 1.2/π
(c) A1 awrt 0.382 cm/s cao
(c) Answer: dh/dt = 6/(5*π) cm/s, awrt 0.382 cm/s
(d) B1 correct statement that dh/dt decreases as h increases
(d) B1 valid justification, e.g. dh/dt is inversely proportional to h² (from dh/dt=120/(π*h²)), so as h grows dh/dt must fall, consistent with the tank widening so the same inflow rate produces a slower rise in depth
(d) Answer: dh/dt decreases as h increases