Core Pure: Polar Coordinates and Hyperbolic Functions Depth
Polar Coordinates and Hyperbolic Functions Depth extends two areas of Core Pure: polar coordinates and hyperbolic functions. In polar coordinates it covers finding the area enclosed by a single loop of a curve that passes through the pole, and the area common to two intersecting polar curves, using A = (1/2) * integral of r^2 dtheta with limits found by solving for where a curve meets the pole or where two curves intersect. In hyperbolic functions it covers deriving and using the logarithmic forms of arsinh, arcosh and artanh, differentiating inverse hyperbolic functions, and applying Osborn's rule to convert trigonometric identities into hyperbolic ones.
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Method
- To find the area enclosed by one loop of a polar curve that passes through the pole, solve r=0 to find the two values of theta that bound that loop, then apply A = (1/2) * integral of r^2 dtheta between those two values.
- To find the area common to (inside both of) two polar curves, find the angle(s) where the curves intersect by solving r_1(theta) = r_2(theta), decide which curve gives the smaller r on each interval of theta, and integrate (1/2)*r^2 using the smaller r on each interval, using symmetry to halve the working where possible.
- Derive the logarithmic form of an inverse hyperbolic function by setting y = arsinh(x) (so x = sinh y), writing sinh y in exponential form, and solving the resulting quadratic in e^y, rejecting the root that cannot be positive, giving arsinh(x) = ln(x+sqrt(x^2+1)).
- Recall, or derive similarly, arcosh(x) = ln(x+sqrt(x^2-1)) for x >= 1, and artanh(x) = (1/2)*ln((1+x)/(1-x)) for -1 < x < 1.
- Differentiate inverse hyperbolic functions either from the log form using the chain rule, or by implicit differentiation on x = sinh y (or x = cosh y), giving standard results such as d/dx[arsinh(x)] = 1/sqrt(x^2+1).
- Use Osborn's rule to convert a known trigonometric identity into the corresponding hyperbolic identity: replace each trig function with its hyperbolic equivalent, and change the sign of any term that is (implicitly or explicitly) a product of two sines.
- Solve equations or prove identities combining several hyperbolic functions by writing everything in terms of cosh and sinh (or in exponential form), then using cosh^2(x) - sinh^2(x) = 1 to reduce to a single-variable equation.
Worked example
The curve C has polar equation r = 1 + 2cos(theta). Find the exact area of the smaller, inner loop of C.
- Solve r=0: 1+2cos(theta)=0, so cos(theta)=-1/2, giving theta = 2*pi/3 or theta = -2*pi/3 (equivalently 4*pi/3), for -pi < theta <= pi.
- The inner loop is traced as theta goes from 2*pi/3 to 4*pi/3 (where r is negative or zero), so Area = (1/2) * integral from theta=2*pi/3 to theta=4*pi/3 of (1+2cos theta)^2 dtheta.
- Expand and simplify the integrand using cos^2(theta) = (1+cos(2theta))/2: (1+2cos theta)^2 = 1+4cos theta+4cos^2theta = 3+4cos(theta)+2cos(2theta).
- Integrate: integral of (3+4cos theta+2cos(2theta)) dtheta = 3theta + 4sin(theta) + sin(2theta).
- Evaluate between the limits: at theta=4*pi/3 the value is 4*pi-(3/2)*sqrt(3); at theta=2*pi/3 the value is 2*pi+(3/2)*sqrt(3); the difference is 2*pi-3*sqrt(3).
- Final answer: Area = (1/2)*(2*pi-3*sqrt(3)) = pi - (3/2)*sqrt(3) (equivalently pi - (3*sqrt(3))/2).
Practice questions
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Q1The curve C has polar equation r = 2cos(theta) - 1. Find the values of theta, in the range -pi < theta <= pi, for which r = 0.Show answer
Answer: theta = pi/3 or theta = -pi/3 (from cos(theta) = 1/2).
Q2Two curves have polar equations r=2 (a circle centred on the pole) and r=4cos(theta) (a circle through the pole). Find the angle theta, with 0 <= theta < pi/2, at which the curves intersect.Show answer
Answer: theta = pi/3 (from 2 = 4cos(theta), so cos(theta) = 1/2).
Q3Explain how you would find the area of the region that lies inside both curves r=f(theta) and r=g(theta), between two points where they intersect.Show answer
Answer: Find the angle(s) of intersection by solving f(theta)=g(theta); between these angles, identify which function gives the smaller r-value (this curve bounds the shared region); integrate (1/2)*r^2, using that smaller r, between the intersection angles, combining more than one interval by symmetry if the region has more than one boundary arc.
Q4Starting from x = sinh(y) = (e^y - e^(-y))/2, show that e^y = x + sqrt(x^2+1), and hence that arsinh(x) = ln(x + sqrt(x^2+1)).Show answer
Answer: Multiplying 2x = e^y - e^(-y) by e^y gives e^(2y) - 2x*e^y - 1 = 0, a quadratic in e^y with solutions e^y = x +- sqrt(x^2+1); since e^y > 0 and sqrt(x^2+1) > |x| always, only e^y = x+sqrt(x^2+1) is valid, so y = arsinh(x) = ln(x+sqrt(x^2+1)).
Q5Differentiate y = arsinh(3x).Show answer
Answer: dy/dx = 3/sqrt(9x^2+1) (using d/dx[arsinh(u)] = u'/sqrt(u^2+1) with u=3x).
Q6State the identity obtained by applying Osborn's rule to the trigonometric identity cos(2theta) = 1 - 2sin^2(theta).Show answer
Answer: cosh(2theta) = 1 + 2sinh^2(theta) (the sign of the sin^2 term flips, since sin^2 is a product of two sines).
Q7Find the exact value of artanh(1/2), giving your answer as a single logarithm.Show answer
Answer: (1/2)*ln(3) (using artanh(x) = (1/2)*ln((1+x)/(1-x)) with x=1/2, giving (1/2)*ln(1.5/0.5) = (1/2)*ln(3)).
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
Two curves have polar equations r=2 and r=4cos(theta). (a) Find the angle theta, with 0 <= theta < pi/2, at which the curves intersect. (b) Find the exact area of the region that lies inside both curves.
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Let y = arcosh(x) for x >= 1, so x = cosh(y) with y >= 0. (a) Show that arcosh(x) = ln(x + sqrt(x^2-1)). (b) Hence find the exact value of arcosh(5/3), giving your answer as a single logarithm in its simplest form.
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(a) Use Osborn's rule to write down the hyperbolic identity corresponding to the trigonometric identity sin(A+B) = sin(A)cos(B) + cos(A)sin(B). (b) Given that sinh(x) = 3 and cosh(x) = sqrt(10), find the exact value of sinh(x + ln 2).
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