Answer to Question #124219 in Calculus for desmond

Question

Suppose a particle P is moving in the plane so that its coordinates are given by P(x,y), where x = 4cos2t, y = 7sin2t.

(i) By finding a,b ∈ R such that x2 a2 + y2 b2 = 1, show that P is travelling on an elliptical path.

(ii) Let L(t) be the distance from P to the origin. Obtain an expression for L(t).

(iii) How fast is the distance between P and the origin changing when t = π/8?[

Answer 1

$x = 4cos\ 2t; y = 7sin\ 2t\implies \frac{x}{4}=cos \ 2t...(1)$$\frac{y}{7}=sin\ 2t...(2)$

(i) Squaring and adding equation (1) and (2), we get

$(\frac{x}{4})^2+(\frac{y}{7})^2=cos^2\ 2t+sin^2\ 2t=1$

So the equation becomes $(\frac{x^2}{4^2})+(\frac{y^2}{7^2})=1$

Hence,$a=4;b=7.$

(ii) $L(t)=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$

$L(t)=\sqrt{(4cos\ 2t)^2+(7sin\ 2t)^2}$ $=\sqrt{16cos^2\ 2t+49sin^2\ 2t}$

$L(t)=\sqrt{16+(49-16)sin^2 \ 2t}=\sqrt{16+33sin^2\ 2t}$

(iii) $\frac{dL}{dt}=$ $\frac{d}{dt}($ $\sqrt{16+33sin^2\ 2t})$ $=\frac{1}{2}\frac{1}{\sqrt{16+33sin^2\ 2t}}33\times 2sin\ 2t\ cos \ 2t\times 2$ $=\frac{33}{\sqrt{16+33sin^2\ 2t}}sin\ 4t$

On putting $t=\frac{\pi}{8}$ ,

$\frac{dL}{dt}=$ $\frac{33}{\sqrt{16+33sin^2\ \frac{2\pi}{8}}}sin\ \frac{4\pi}{8}=$ $\frac{33}{\sqrt{16+33sin^2\ \frac{\pi}{4}}}sin\ \frac{\pi}{2}=$ $\frac{33}{\sqrt{16+33(\frac{1}{\sqrt{2}})^2}}=5.79$