Question

if x=e^t, y=sin t, find d^2x/dx^2 when t=π/2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x=e^{t}\,\,\,\,\&\,\,\,\,y=sin(t)}$

Differentiating each term w.r.t t,

$\sf{\dfrac{dx}{dt}=e^{t}\,\,\,\,\&\,\,\,\,\dfrac{dy}{dt}=cos(t)}$

So,

$\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{cos(t)}{e^{t}}}$

$\sf{\implies\dfrac{dy}{dx}=e^{-t}\,cos(t)}$

$\sf{\implies\dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg)=\dfrac{d}{dx}\bigg(e^{-t}\,cos(t)\bigg)}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=\dfrac{d}{dt}\bigg(e^{-t}\,cos(t)\bigg)\cdot\dfrac{dt}{dx}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=\bigg\{cos(t)\cdot\dfrac{d}{dt}(e^{-t})+e^{-t}\cdot\dfrac{d}{dt}(cos(t))\bigg\}\cdot\dfrac{dt}{dx}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=\bigg\{-cos(t)\cdot\,e^{-t}-e^{-t}\cdot\,sin(t)\bigg\}\cdot\dfrac{dt}{dx}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=-e^{-t}\{cos(t)+sin(t)\}\cdot\dfrac{dt}{dx}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=-e^{-t}\{cos(t)+sin(t)\}\cdot\dfrac{1}{e^{t}}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}=-e^{-2t}\{cos(t)+sin(t)\}}$

Now,

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}\bigg|_{t=\frac{\pi}{2}}=-e^{-\pi}\bigg\{cos\bigg(\dfrac{\pi}{2}\bigg)+sin\bigg(\dfrac{\pi}{2}\bigg)\bigg\}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}\bigg|_{t=\frac{\pi}{2}}=-e^{-\pi}\{0+1\}}$

$\sf{\implies\dfrac{d^{2}y}{dx^{2}}\bigg|_{t=\frac{\pi}{2}}=-e^{-\pi}}$