If the solve the problem

Question:

If $x=3 \cot t-2 \cos ^{3} t, y=3 \sin t-2 \sin ^{3} t$, find $\frac{d^{2} y}{d x^{2}}$

Solution:

Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=y_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{X}$

(iv) $\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$

(v) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

(vi) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

(vii) parameteric forms $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

given: -

$x=3 \cot t-2 \cos ^{3} t, y=3 \sin t-2 \sin ^{3} t$

differentiating both w.r.t t

$\frac{\mathrm{dx}}{\mathrm{dt}}=-3 \sin \mathrm{t}-6 \cos ^{2} \mathrm{t}(-\sin \mathrm{t})$

$\frac{\mathrm{dx}}{\mathrm{dt}}=-3 \sin \mathrm{t}+6 \cos ^{2} \mathrm{tsin} \mathrm{t}$

And $y=3 \sin t-2 \sin ^{3} t$

differentiating both w.r.t t

$\frac{d y}{d t}=3 \cos t-6 \sin ^{2} t \cos t$

Now,

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos t-2 \sin ^{2} t \cos t}{-\sin t+2 \cos ^{2} t \sin t}$

$\Rightarrow \frac{d y}{d x}=\frac{\operatorname{cost}\left[1-2 \sin ^{2} t\right]}{\sin t\left[2 \cos ^{2} t-1\right]}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cot \mathrm{t}$

differentiating both w.r.t $x$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$

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