Question

Solve C and D part please

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter.(Can you name the curve?) (a) r = sin 0, y = cos, - SOST (b) x = 4 sect and y = 3 tant 3 3 (c) r=-1+ z sint and y = cost for -A <t <3 (d) x = cosht and y cosh 3t (no need to sketch this curve) 1 2

Answer 1

(c)

$\small x=-1+\frac{3}{2}\sin t;\; y=\frac{1}{2}-\frac{3}{2}\cos t; \; -\pi \leq t \leq 3\pi$

Firstly, let us calculate the values of x and y for

$\small t=-\pi,-\frac{\pi}{2},0,\frac{\pi}{2},\pi$

we have respectively the points (2,-1),(-2.5,0.5),(-1,1),(0.5,0.5),(2,-1).

Thus, we can understand two things from this:

1. As t increases, the curve traces a path traveling counterclockwise through these points.

2. After coming back to (2,-1), the curve traces through the same points for other multiples of pi

Thus, we can form some form of abstract cycling shape now. Now, let us eliminate the parameter.

$\small x=-1+\frac{3}{2}\sin t;\; y=\frac{1}{2}-\frac{3}{2}\cos t \\ \\ \Rightarrow x+1=\frac{3}{2}\sin t; \; y-\frac{1}{2}=-\frac{3}{2}\cos t \text{ [squaring and adding]} \\ \\ \Rightarrow (x+1)^2+\left (y-\frac{1}{2} \right )^2=\frac{9}{4}=\left (\frac{3}{2} \right )^2;\;\;\;\, \mathbf{ [\because sin^2t+cos^2t=1]}$

which is the equation of a circle with centre at (-1,0.5) and radius = 1.5, and orientation = counterclockwise, as t increases.

(d)

We are given

$\small x=\cosh t=\frac{e^t+e^{-t}}{2} ; \;\; y=\cosh 3t=\frac{e^3^t+e^{-3t}}{2} \\ \Rightarrow 2x=e^t+e^{-t}; 2y=e^3^t+e^{-3t}\Rightarrow (2x)^3=e^{3t}+e^{-3t}+3.e^t.e^{-t}(e^t+e^{-t}) \\ \Rightarrow 8x^3=y+3.2x \Rightarrow 8x^3=y+6x$

which is our required expression after eliminating the parameter.