calculusderivativesfunctionsmaxima-minima
Question:
Find the maximum distance from origin of a point on the curve $x=a\sin t-b\sin\left(\frac{at}b\right), y=a\cos t-b\cos\left(\frac{at}b\right)$, both $a,b\gt0$
My Attempt:
Applying distance formula,
$D^2=a^2+b^2-2ab\cos(t-\frac{at}b)$
For maximum distance, cosine should be minimum i.e. $-1$. Thus, maximum distance is $a+b$.
But the answer given is $a-b$.
Am I missing anything?
We have $$\eqalign{I_m(a,b)= \int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}\,\mathrm dx &=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\\\cr \frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0 }\right.}$$ Proof can be seen here. Hence \begin{align} \int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx\\ &=\frac{1}{2}\left[\frac{\pi}{a^2-b^2}-\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^2\right]\\ &=\frac{\pi}{2 a^2} \end{align}
The curve is an involute of circle with radius $a$.
Tangent of the curve: \begin{align*} \frac{dx}{d\theta} &= -a\sin \theta+a\theta \cos \theta+a\sin \theta \\ &= a\theta \cos \theta \\ \frac{dy}{d\theta} &= a\cos \theta+a\theta \sin \theta-a\cos \theta \\ &= a\theta \sin \theta \\[5pt] \frac{dy}{dx} &= \frac{a\theta \sin \theta}{a\theta \cos \theta} \\[5pt] &= \tan \theta \\ \end{align*}
Equation of the Normal: \begin{align*} y-(a\sin \theta-a\theta \cos \theta) &= -\cot \theta [x-(a\sin \theta+a\theta \cos \theta)] \\ y\sin \theta -a\sin \theta(\sin \theta-\theta \cos \theta) &= -x\cos \theta +a\cos \theta(\cos \theta+\theta \sin \theta) \\ x\cos \theta+y\sin \theta &= a (\sin^2 \theta+\cos^2 \theta) \\ x\cos \theta+y\sin \theta &= a \end{align*} which has fixed distance $a$ from the origin.
Best Answer
Let $D(t)$ be the distance of point P from origi On. Then $$OP=D(t)=\sqrt{a^2+b^2-2ab\cos c t}, c=(b-a)/a,~ a,b>0$$ $D(t)$ will admit maximum value when $ct=\pi$, hence $D_{max}=a+b.$