Question

f(x,y)=sin x+cos y+xy^2 ,x=cos t,y=sin t find df/dt at t=π/2

Answer 1

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Answer 2

Answer:

The derivative $\frac{df}{dt}$ at $t= \frac{\pi}{2}$ is -2.

Step-by-step explanation:

Given the function on two variables,

$f(x,y) = \sin x+\cos y+xy^2$

Also

$x= \cos t\\\\y=\sin t$

We need to find   $\frac{df}{dt} \text{ at } t= \frac{\pi}{2}$

Here we have to use the chain rule. Since $f$ is a function of $x$ and $y$ and $x \text{ and } y$ functions on $t$ , the chain rule gives,

$\dfrac{df}{dt} = \dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+ \dfrac{\partial f}{\partial y}\dfrac{dy}{dt}$

So we need to find the four terms on the right-hand side. When we find the partial derivative with respect to  $x$, we consider $y$  as a constant and vice versa.

$\dfrac{\partial f}{\partial x} = \dfrac{\partial }{\partial x}( \sin x+\cos y+xy^2) = \cos x +y^2\\\\\\\dfrac{\partial f}{\partial y} = \dfrac{\partial }{\partial y}( \sin x+\cos y+xy^2) =-\sin y+2yx\\\\\\\dfrac{dx}{dt} = \dfrac{d}{dt}\cos t = -\sin t \\\\\\\dfrac{dy}{dt} = \dfrac{d}{dt}\sin t = \cos t$

Thus we get,

$\dfrac{df}{dt} = (\cos x+y^2)(-\sin t)+(-\sin y+2xy)\cos t\\ \\=(\cos(\cos t)+ \sin^2t) (-\sin(t))+(-\sin(\sin t)+2\cos t\sin t)\cos t$

When $t=\frac{\pi}{2}$ ,

$cos \frac{\pi}{2} =0\\\\\sin \frac{\pi}{2} =1$

Hence

$\dfrac{df}{dt} = (\cos( \cos\frac{\pi}{2} )+\sin^ \frac{\pi}{2})(-\sin \frac{\pi}{2})=(\cos(0)+1)(-1) =(1+1)(-1)=-2$