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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 Marks
Question
Let C be a curve defined parametrically as $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3\theta,0\leq\theta\leq\frac{\pi}{2}.$ Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).
✓

Answer

As, $\text{x}=\text{a}\cos^3\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta$
And, $\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin^2\theta\cos\theta$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}=-\tan\theta$
For the tangent to be parallel to the chord joining the points (a, 0) and (0, a).
$\frac{\text{dy}}{\text{dx}}=\frac{0-\text{a}}{\text{a}-0}$
$\Rightarrow-\tan\theta=-1$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
Now,
$\text{x}=\text{a}\cos^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$ and
$\text{y}=\text{a}\sin^3\frac{\pi}{4}=\text{a}\Big(\frac{1}{\sqrt{2}}\Big)^3=\frac{\text{a}}{2\sqrt2}$
So, the point P on the curve C is $\Big(\frac{\text{a}}{2\sqrt2},\frac{\text{a}}{2\sqrt2}\Big).$

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