Question

If x= a (cost + tsint) y = a(sint-tcost) find d2y/dx2

Answer 1

Answer:

$\frac{d^{2}y }{dx^{2}}=\frac{sec^{3}t}{at}$

Step-by-step explanation:

$x=a(cost+tsint)$

Differentiating this equation with respect to t, we have

⇒$\frac{dx}{dt}=a(-sint)+a(tcost+sint)$

⇒$\frac{dx}{dt}=atcost$                                                                

Now, $y=a(sint-tcost)$

Differentiating with respect to t, we have

⇒$\frac{dy}{dt}=acost-a(-tsint+cost)$

⇒$\frac{dy}{dt}=atsint$

Now, $\frac{dy}{dx}=\frac{dy}{dt}{\times}\frac{dt}{dx}$

⇒$\frac{dy}{dx}=(atsint){\times}\frac{1}{(atcost)}$

⇒$\frac{dy}{dx}=tant$

Again differentiating this equation with respect to x, we have

⇒$\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}(tant)$

⇒$\frac{d^{2}y }{dx^{2}}=\frac{d}{dx}(tant)\frac{dt}{dx}$

Now, substituting the value of $\frac{dt}{dx}$  in above equation, we have

⇒$\frac{d^{2}y }{dx^{2}}=sec^{2}t{\times}\frac{1}{atcost}$

⇒$\frac{d^{2}y }{dx^{2}}=\frac{sec^{2}t}{at{\times}\frac{1}{sect}}$

⇒$\frac{d^{2}y }{dx^{2}}=\frac{sec^{3}t}{at}$