Let $z = e ^ { x ^ { 2 } } \sin y$ , and let $x$ and $y$ be functions of $t$ with $x ( 1 ) = 0 , y ( 1 ) = 0 , x ^ { \prime } ( 1 ) = 3$ , and $y ^ { \prime } ( 1 ) = 4$ . Find $d z / d t$ when $t = 1$ .
A)0
B)1
C)2
D)3
E)4
F)5
G)6
H)7

Question

Quizplus · Accepted Answer

To find $dz/dt$ when $t = 1$, we use the chain rule for differentiation. Given $z = e^{x^2} \sin y$, we have:$$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$First, we find the partial derivatives:$$\frac{\partial z}{\partial x} = 2x e^{x^2} \sin y$$$$\frac{\partial z}{\partial y} = e^{x^2} \cos y$$At $t = 1$, $x(1) = 0$, $y(1) = 0$, $x'(1) = 3$, and $y'(1) = 4$. Substituting these values:$$\frac{dz}{dt} = (2 \cdot 0 \cdot e^{0^2} \cdot \sin 0) \cdot 3 + (e^{0^2} \cdot \cos 0) \cdot 4 = 0 + 4 = 4$$Therefore, $dz/dt = 4$ when $t = 1$.

Let z=ex2sin⁡yz = e ^ { x ^ { 2 } } \sin yz=ex2siny

Let $z = e ^ { x ^ { 2 } } \sin y$