If x = a and y = a ( + t 2 ), find d^ 2 y dx^ 2 .

Question

If $\text{x} = \text{a}\sin\text{t} \text{ and } \text{y = a} \bigg(\cos\text{t} + \log\tan\frac{\text{t}}{2}\bigg),\text{ find } \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}.$

✓

Answer

Here, x = a sin t, y = a $\bigg[\cos\text{t} + \log\bigg(\tan\frac{\text{t}}{2}\bigg)\bigg]$
$\because$ x = a sin t
Differentiating both sides w.r.t. t, we get
$\frac{\text{dx}}{\text{dt}}= \text{a}\cos\text{t}$ - - - - - (i)
Again, $\because\text{y = a }\bigg[\cos\text{t} + \log\bigg(\tan\frac{\text{t}}{2}\bigg)\bigg]$
Differentiating both sides w.r.t. t we get
$\frac{\text{dy}}{\text{dt}} = \text{a}\bigg[-\sin\text{t} + \frac{1}{\tan\frac{\text{t}}{2}}.\sec^{2}\frac{\text{t}}{2}.\frac{\text{t}}{2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}} = \text{a}\bigg[ -\sin\text{t} + \frac{1}{\sin\text{t}}\bigg]\Rightarrow\frac{\text{dy}}{\text{dt}} = \frac{\text{a}(1 - \sin^{2}\text{t})}{\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}} = \frac{\text{a}\cos^{2}\text{t}}{\sin\text{t}}$ - - - - - -(ii)
$\because\frac{\text{dy}}{\text{dx}} =\frac{\text{dy}/\text{dt}}{\text{dx}/\text{dt}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}} =\frac{\text{a}\cos^{2}\text{t}}{\sin\text{t}}\times\frac{1}{\text{a}\cos\text{t}}$ [From (i) and (ii)]
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \cot\text{t}$
Differentiating again w.r.t. x we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = 0-\text{cosec}^{2}\text{t}.\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = -\text{cosec}^{2}\text{t}.\frac{1}{\text{a}\cos\text{t}} = \frac{-\text{cosec}^{2}\text{t}}{\text{a}\cos\text{t}}.$