Revision Library

revisionlibrary.co.uk

A-Level · Further Pure Mathematics 1 and 2 (Polar Coordinates and Hyperbolic Functions Depth)

FP.CP15 Core Pure: Polar Coordinates and Hyperbolic Functions

EDEXCEL 9FM0 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

You are given that sinh(A+B) = sinh(A)cosh(B) + cosh(A)sinh(B) and cosh(A+B) = cosh(A)cosh(B) + sinh(A)sinh(B) for all real A and B.

(a)Prove that tanh(A+B) = (tanh(A) + tanh(B)) / (1 + tanh(A)tanh(B)).(5)

(b)Given that tanh(x) = 1/3, use the result from part (a) to find the exact value of tanh(2x).(4)

(Total for Question 1 is 9 marks)

This question concerns the auxiliary form of expressions such as R cosh(x + a).

(a)Show that 5cosh(x) + 3sinh(x) can be written in the form 4cosh(x + ln(2)).(6)

(b)Hence solve, for x in the real numbers, the equation 5cosh(x) + 3sinh(x) = 4sqrt(2), giving each solution as an exact logarithm.(6)

(Total for Question 2 is 12 marks)

This question concerns the inverse hyperbolic function arcosh.

(a)Given that y = arcosh(x) for x ≥ 1, y ≥ 0, so that x = cosh(y), show that arcosh(x) = ln(x + √x² - 1).(5)

(b)Hence find the exact value of arcosh(13/5), giving your answer as a single logarithm.(3)

(c)Hence, by differentiating the result of part (a) directly with respect to x, show that d/dx(arcosh(x)) = 1/√x² - 1, for x > 1.(4)

(Total for Question 3 is 12 marks)

This question concerns the inverse hyperbolic function artanh.

(a)Given that y = artanh(x) for |x| < 1, so that x = tanh(y), show that artanh(x) = (1/2)ln((1+x)/(1-x)).(5)

(b)Hence solve, for x in the real numbers, the equation 3tanh(x) = 2, giving your answer as an exact single logarithm.(3)

(c)Differentiate artanh(2x) with respect to x, simplifying your answer.(3)

(Total for Question 4 is 11 marks)

This question tests differentiation and integration of hyperbolic functions.

(a)Differentiate cosh(1 - 4x²) with respect to x.(2)

(b)Differentiate ln(sinh(3x)) with respect to x, for x > 0, simplifying your answer.(3)

(c)Differentiate x² tanh(x) with respect to x.(3)

(d)Find, in simplest form, the integral of x cosh(x²) with respect to x.(2)

(e)Find the integral of tanh(x) with respect to x, giving your answer in the form ln(f(x)) + c.(3)

(Total for Question 5 is 13 marks)

This question uses hyperbolic substitutions to derive and apply standard integration results.

(a)Using the substitution x = 3cosh(u), show that, for x > 3, the integral of 1/√x² - 9 with respect to x is arcosh(x/3) + c.(5)

(b)Hence find the exact value of the integral, from x=4 to x=6, of 1/√x²-9 with respect to x, giving your answer in terms of logarithms.(4)

(c)Using the substitution x = 5tanh(u), show that, for -5 < x < 5, the integral of 1/(25 - x²) with respect to x is (1/5)artanh(x/5) + c.(5)

(d)Hence find the exact value of the integral, from x=0 to x=3, of 1/(25 - x²) with respect to x, giving your answer as a single natural logarithm.(4)

(Total for Question 6 is 18 marks)

The curve C has polar equation r = 6(θ), for θ ≥ 0 (an Archimedean spiral). The point P lies on C where θ = π/3.

(a)Find the exact Cartesian coordinates of P.(4)

(b)Hence write down the exact distance OP, where O is the pole.(1)

(Total for Question 7 is 5 marks)

A curve has Cartesian equation (x-1)² + (y-1)² = 2.

(a)Show that the polar equation of this curve can be written as r = 2cos(θ) + 2sin(θ), stating the range of values of θ, 0 ≤ θ < 2pi, for which r ≥ 0.(5)

(b)Hence write down the maximum value of r on the curve, and state the value of θ, 0 ≤ θ < 2pi, at which this maximum occurs.(3)

(Total for Question 8 is 8 marks)

The curve C has polar equation r² = 12cos(2theta), for -π/4 ≤ θ ≤ π/4 (one loop of a lemniscate).

(a)Write down the values of θ, in the given interval, for which r = 0.(1)

(b)Show that the area of the loop of C described is 6.(5)

(c)Write down the total area enclosed by both loops of the full curve r² = 12cos(2theta), justifying your answer briefly.(2)

(Total for Question 9 is 8 marks)

The curve C has polar equation r = 4 + 4cos(θ), for 0 ≤ θ < 2pi.

(a)Find the area enclosed by C.(6)

(b)Find the area of the region that lies inside C and outside the circle with polar equation r = 4.(7)

(Total for Question 10 is 13 marks)

The curve C has polar equation r = 2 + 2sin(θ), for 0 ≤ θ < 2pi.

(a)Show that dy/dtheta = 2cos(θ)(1 + 2sin(θ)), where y = r*sin(θ).(3)

(b)Find, in exact form, the values of θ in the interval 0 ≤ θ < 2pi for which the tangent to C is parallel to the initial line, explaining why θ = 3pi/2 must be rejected.(6)

(Total for Question 11 is 9 marks)

The curve C has polar equation r = 1 + 3cos(θ), for 0 ≤ θ < 2pi. Since the coefficient of cos(θ) (3) is greater than the constant term (1), C has an inner loop.

(a)Find, to 3 significant figures, the values of θ in the interval 0 ≤ θ < 2pi for which r = 0.(2)

(b)Show that dx/dtheta = -sin(θ)(1 + 6cos(θ)), where x = r*cos(θ), and hence find, to 3 significant figures, the values of θ in the interval 0 ≤ θ < π for which the tangent to C is perpendicular to the initial line.(8)

(Total for Question 12 is 10 marks)

Mark scheme · FP.CP15 Core Pure: Polar Coordinates and Hyperbolic Functions

Question 1

(a) M1 write tanh(A+B) = sinh(A+B)/cosh(A+B) and substitute the given addition formulae
(a) M1 divide every term in the numerator and denominator by cosh(A)cosh(B)
(a) A1 numerator simplifies to tanh(A) + tanh(B), using sinh(A)/cosh(A) = tanh(A)
(a) A1 denominator simplifies to 1 + tanh(A)tanh(B), using sinh(A)sinh(B)/(cosh(A)cosh(B)) = tanh(A)tanh(B)
(a) A1 cso: tanh(A+B) = (tanh(A)+tanh(B))/(1+tanh(A)tanh(B))
(a) Answer: tanh(A+B) = (tanh(A)+tanh(B)) / (1+tanh(A)tanh(B)), proved.
(b) M1 use tanh(2x) = 2tanh(x)/(1+tanh²(x)), the case A=B=x of part (a)
(b) M1 substitute tanh(x) = 1/3
(b) A1 simplify (2/3)/(1+1/9) = (2/3)/(10/9)
(b) A1 cao: tanh(2x) = 3/5
(b) Answer: tanh(2x) = 3/5

Question 2

(a) M1 write R cosh(x+a) = R cosh(a)cosh(x) + R sinh(a)sinh(x) and match coefficients: R cosh(a) = 5, R sinh(a) = 3
(a) M1 add and subtract the two equations, using e^a = cosh(a)+sinh(a), to get R e^a = 8 and R e^-a = 2
(a) M1 multiply these two results: R² = 16
(a) A1 R = 4 (R > 0)
(a) M1 divide the two results: e^2a = 4, so e^a = 2 (e^a > 0)
(a) A1 cso: a = ln(2), giving 5cosh(x)+3sinh(x) = 4cosh(x+ln(2))
(a) Answer: 5cosh(x) + 3sinh(x) = 4cosh(x + ln(2))
(b) M1 use the result of part (a) to write 4cosh(x+ln(2)) = 4sqrt(2)
(b) M1 cosh(x+ln(2)) = √2
(b) M1 use arcosh(√2) = ln(√2 + √2-1) = ln(√2+1), so x+ln(2) = +-ln(√2+1)
(b) A1 both values of x+ln(2) correctly identified
(b) M1 rearrange each to make x the subject: x = ln(√2+1) - ln(2) or x = -ln(√2+1) - ln(2)
(b) A1 cao: x = ln((√2+1)/2) or x = -ln(2sqrt(2)+2) (oe unsimplified forms accepted)
(b) Answer: x = ln((√2+1)/2) or x = -ln(2sqrt(2)+2)

Question 3

(a) M1 state x = cosh(y)
(a) M1 use cosh²(y) - sinh²(y) = 1 to write sinh(y) = √x² - 1
(a) A1 justify the positive square root, since y ≥ 0 implies sinh(y) ≥ 0
(a) M1 use e^y = cosh(y) + sinh(y) = x + √x² - 1
(a) A1 cso: y = ln(x + √x²-1), so arcosh(x) = ln(x + √x²-1)
(a) Answer: arcosh(x) = ln(x + √x²-1)
(b) M1 substitute x = 13/5 into ln(x + √x²-1)
(b) A1 simplify √169/25 - 1 = √144/25 = 12/5
(b) A1 cao: arcosh(13/5) = ln(5)
(b) Answer: arcosh(13/5) = ln(5)
(c) M1 differentiate ln(x+√x²-1) using the chain rule: (1/(x+√x²-1)) * (1 + x/√x²-1)
(c) M1 combine the bracket over a common denominator: (√x²-1+x)/√x²-1
(c) A1 cancel the common factor (x+√x²-1) from numerator and denominator
(c) A1 cso: d/dx(arcosh(x)) = 1/√x²-1
(c) Answer: d/dx(arcosh(x)) = 1/√x²-1, for x > 1

Question 4

(a) M1 write x = tanh(y) = (e^2y-1)/(e^2y+1), obtained by multiplying numerator and denominator of (e^y-e^-y)/(e^y+e^-y) by e^y
(a) M1 form the equation x(e^2y+1) = e^2y - 1
(a) M1 rearrange to isolate e^2y: x + 1 = e^2y(1-x)
(a) A1 e^2y = (1+x)/(1-x)
(a) A1 cso: y = (1/2)ln((1+x)/(1-x))
(a) Answer: artanh(x) = (1/2)ln((1+x)/(1-x))
(b) M1 tanh(x) = 2/3, substituted into x = (1/2)ln((1+x)/(1-x)) with x=2/3
(b) A1 simplify (1+2/3)/(1-2/3) = (5/3)/(1/3) = 5
(b) A1 cao: x = (1/2)ln(5) (= ln(√5))
(b) Answer: x = (1/2)ln(5) = ln(√5)
(c) M1 use d/dx[artanh(u)] = u'/(1-u²) with u = 2x
(c) A1 correct substitution u=2x, u'=2
(c) A1 cao: d/dx[artanh(2x)] = 2/(1-4x²)
(c) Answer: d/dx[artanh(2x)] = 2/(1-4x²)

Question 5

(a) M1 apply the chain rule: derivative of cosh(u) is sinh(u) times u', with u = 1-4x², u' = -8x
(a) A1 cao: -8x sinh(1-4x²)
(a) Answer: -8x sinh(1 - 4x²)
(b) M1 apply the chain rule: (1/sinh(3x)) times derivative of sinh(3x)
(b) A1 derivative of sinh(3x) = 3cosh(3x) used correctly
(b) A1 cao: simplify (3cosh(3x))/sinh(3x) = 3coth(3x)
(b) Answer: 3coth(3x)
(c) M1 apply the product rule u'v + uv' with u = x², v = tanh(x)
(c) A1 use derivative of tanh(x) = sech²(x) correctly
(c) A1 cao: 2x tanh(x) + x² sech²(x)
(c) Answer: 2x tanh(x) + x² sech²(x)
(d) M1 recognise x cosh(x²) as (1/2) times the derivative of sinh(x²)
(d) A1 cao: (1/2)sinh(x²) + c
(d) Answer: (1/2)sinh(x²) + c
(e) M1 write tanh(x) = sinh(x)/cosh(x) and recognise the numerator as the derivative of the denominator
(e) A1 integral form ln|cosh(x)| + c
(e) A1 cao: since cosh(x) > 0 for all real x, this simplifies to ln(cosh(x)) + c
(e) Answer: ln(cosh(x)) + c

Question 6

(a) M1 substitute x = 3cosh(u), dx = 3sinh(u) du
(a) M1 use cosh²(u) - 1 = sinh²(u) to write √x²-9 = √9cosh²(u)-9 = 3sinh(u)
(a) A1 integral simplifies to the integral of 1 with respect to u
(a) M1 integrate to get u + c and use u = arcosh(x/3)
(a) A1 cso: integral of 1/√x²-9 dx = arcosh(x/3) + c
(a) Answer: arcosh(x/3) + c
(b) M1 substitute the limits into arcosh(x/3): arcosh(2) - arcosh(4/3)
(b) M1 use arcosh(2) = ln(2 + √4-1) = ln(2+√3)
(b) A1 use arcosh(4/3) = ln(4/3 + √16/9-1) = ln((4+√7)/3)
(b) A1 cao: value = ln(2+√3) - ln((4+√7)/3), oe (e.g. a single combined logarithm)
(b) Answer: ln(2+√3) - ln((4+√7)/3), approx 0.522 (3 s.f.)
(c) M1 substitute x = 5tanh(u), dx = 5sech²(u) du
(c) M1 use 1 - tanh²(u) = sech²(u) to write 25-x² = 25(1-tanh²(u)) = 25sech²(u)
(c) A1 integral simplifies to the integral of 1/5 with respect to u
(c) M1 integrate to get u/5 + c and use u = artanh(x/5)
(c) A1 cso: integral of 1/(25-x²) dx = (1/5)artanh(x/5) + c
(c) Answer: (1/5)artanh(x/5) + c
(d) M1 substitute the limits into (1/5)artanh(x/5): (1/5)[artanh(3/5) - artanh(0)]
(d) A1 artanh(0) = 0, so the value is (1/5)artanh(3/5)
(d) M1 use artanh(3/5) = (1/2)ln((1+3/5)/(1-3/5)) = (1/2)ln(4) = ln(2)
(d) A1 cao: (1/5)ln(2)
(d) Answer: (1/5)ln(2)

Question 7

(a) M1 find r = 6(π/3) = 2pi at θ = π/3
(a) M1 use the exact values cos(π/3) = 1/2 and sin(π/3) = √3/2
(a) A1 x = r*cos(π/3) = 2pi*(1/2) = π
(a) A1 y = r*sin(π/3) = 2pi*(√3/2) = π*√3
(a) Answer: P = (π, π*√3)
(b) B1 OP = r = 2pi (the distance from the pole is simply the value of r at that θ)
(b) Answer: OP = 2pi

Question 8

(a) M1 expand (x-1)² + (y-1)² = 2 to x² + y² - 2x - 2y = 0
(a) M1 substitute x²+y² = r², x = r*cos(θ), y = r*sin(θ): r² - 2r*cos(θ) - 2r*sin(θ) = 0
(a) M1 divide by r (r not identically 0): r = 2cos(θ) + 2sin(θ), cso
(a) A1 find where r = 0: tan(θ) = -1, giving θ = 3pi/4 or θ = 7pi/4
(a) A1 cao: r ≥ 0 for 0 ≤ θ ≤ 3pi/4 or 7pi/4 ≤ θ < 2pi
(a) Answer: r = 2cos(θ) + 2sin(θ); r ≥ 0 for 0 ≤ θ ≤ 3pi/4 or 7pi/4 ≤ θ < 2pi
(b) M1 express r = 2cos(θ) + 2sin(θ) in the form R*cos(θ - α) with R = 2sqrt(2), tan(α) = 1
(b) M1 identify that the maximum value of r is R, occurring when θ = α = π/4
(b) A1 cao: maximum r = 2sqrt(2), at θ = π/4
(b) Answer: Maximum r = 2sqrt(2), at θ = π/4

Question 9

(a) B1 θ = -π/4 and θ = π/4 (both required)
(a) Answer: θ = -π/4, π/4
(b) M1 Area = (1/2) times the integral of r² with respect to θ, with limits -π/4 to π/4
(b) A1 correct integral: (1/2) times the integral of 12cos(2theta) dtheta
(b) M1 integrate to obtain 3sin(2theta) (or equivalent antiderivative)
(b) M1 substitute the limits: 3[sin(π/2) - sin(-π/2)]
(b) A1 cao: Area = 3[1-(-1)] = 6
(b) Answer: Area = 6 (square units)
(c) B1 recognise that the second loop (for θ between 3pi/4 and 5pi/4) is congruent to the first by symmetry
(c) B1 cao: total area = 2 x 6 = 12
(c) Answer: Total area = 12 (square units)

Question 10

(a) M1 Area = (1/2) times the integral of r² with respect to θ, with limits 0 to 2pi
(a) M1 expand (4+4cos(θ))² = 16 + 32cos(θ) + 16cos²(θ)
(a) M1 use cos²(θ) = (1+cos(2theta))/2 to rewrite the integrand
(a) A1 correct integral: (1/2) times the integral of (16 + 32cos(θ) + 8 + 8cos(2theta)) dtheta
(a) M1 integrate term by term and substitute the limits 0 to 2pi (the cos(θ) and cos(2theta) terms integrate to zero over a full period)
(a) A1 cao: Area = 24pi
(a) Answer: Area = 24pi
(b) M1 find the points of intersection: 4+4cos(θ) = 4 gives cos(θ) = 0, so θ = +-π/2
(b) M1 set up Area = (1/2) times the integral, with limits -π/2 to π/2, of [(4+4cos(θ))² - 4²] dtheta
(b) M1 expand the integrand to 32cos(θ) + 16cos²(θ)
(b) A1 use cos²(θ) = (1+cos(2theta))/2 to write the integrand as 32cos(θ) + 8 + 8cos(2theta)
(b) dM1 integrate to obtain 32sin(θ) + 8theta + 4sin(2theta)
(b) A1 correct evaluation between the limits gives 64 + 8pi
(b) A1 cao: Area = (1/2)(64+8pi) = 32 + 4pi
(b) Answer: Area = 32 + 4pi

Question 11

(a) M1 write y = (2+2sin(θ))sin(θ) = 2sin(θ) + 2sin²(θ)
(a) M1 differentiate to get dy/dtheta = 2cos(θ) + 4sin(θ)cos(θ)
(a) A1 cso: factorise to dy/dtheta = 2cos(θ)(1 + 2sin(θ))
(a) Answer: dy/dtheta = 2cos(θ)(1 + 2sin(θ))
(b) M1 set dy/dtheta = 0
(b) M1 solve cos(θ) = 0 to get θ = π/2 or θ = 3pi/2
(b) M1 solve sin(θ) = -1/2 to get θ = 7pi/6 or θ = 11pi/6
(b) A1 all four values π/2, 3pi/2, 7pi/6, 11pi/6 correctly identified as solutions of dy/dtheta = 0
(b) B1 recognise that r = 2+2sin(θ) = 0 at θ = 3pi/2, so C passes through the pole there and the tangent direction is not given by this stationary-point condition, so θ = 3pi/2 is rejected
(b) A1 cao: θ = π/2, 7pi/6, 11pi/6
(b) Answer: θ = π/2, 7pi/6, 11pi/6

Question 12

(a) B1 solve cos(θ) = -1/3
(a) B1 θ = 1.91 or θ = 4.37 (both awrt 3 s.f.)
(a) Answer: θ = 1.91 or θ = 4.37 (radians, 3 s.f.)
(b) M1 write x = (1+3cos(θ))cos(θ) = cos(θ) + 3cos²(θ)
(b) M1 differentiate to get dx/dtheta = -sin(θ) - 6sin(θ)cos(θ)
(b) A1 cso: factorise to dx/dtheta = -sin(θ)(1 + 6cos(θ))
(b) M1 set dx/dtheta = 0 and split into sin(θ) = 0 or cos(θ) = -1/6
(b) A1 θ = 0 is a solution in the given range (from sin(θ)=0; θ=π is excluded as the interval is 0≤θ<π)
(b) M1 solve cos(θ) = -1/6 using a calculator
(b) A1 θ = 1.74 (3 s.f.)
(b) B1 confirm r is non-zero at both θ=0 (r=4) and θ=1.74 (r=0.5), so both give genuine tangent lines (neither coincides with the pole angles from part (a))
(b) Answer: θ = 0 or θ = 1.74 (3 s.f.)

Core Pure: Polar Coordinates and Hyperbolic Functions - Worksheets, Questions and Revision