Hey first of all, is this actually part of the new study design for Specialist Maths?
If it is, can you please solve the question below? I tried using the formula but I end up with an integral which I cannot solve.
A curve is specified parametrically by the equations
x=t−sin(t), y=1−cos(t) Find the length of the curve from t = 0
to t = 2π.
Thank you
Arc length obviously is, but I'm not sure about whether it is in the study design for parametric equations, it doesn't explicitly say it is (or that it isn't!). The idea is pretty simple for arc length of a curve defined by parametric equations (google the derivation). Anyway if it was in an exam it may be likely they would guide students subtly through how to find the arc length in the case where you have parametric equations.
The integral is solvable in terms of specialist techniques.
s=integral of sqrt((dx/dt)^2+(dy/dt)^2)dt
I won't evaluate the integral (the answer is 8 for when you get to it
) instead I'll show you how you'd make it solveable.
2 steps basically, first foil out the (1-cos(t))^2, second is to use the cosine double angle formula.
So within the square root we have
(1-cos(t))^2+sin(t)^2
=1-2cos(t)+cos(t)^2+sin(t)^2
=2-2cos(t) since cos(t)^2+sin(t)^2=1
=2(1-cos(t))
=2(1-(2cos(t/2)^2-1))
=2-4cos(t/2)^2+2
=4(1-cos(t/2)^2)
=sin(t/2)^2 (you can move that 4 outside of the square root and it becomes 2. Now you have sqrt(sin(t/2)^2), we take the positive root due to the upper and lower limits of the integral, so finally we have 2*integral(sin(t/2))dt from 0 to 2pi
Home
Help
Search
Login
