Question

2 points) Suppose w=xy+yzw=xy+yz, where x=et, y=2+sin(t)x=et, y=2+sin⁡(t), and z=2+cos(3t)z=2+cos⁡(3t). A ) Use the chain rule to find dwdtdwdt as a function of x, y, z, and t. Do not rewrite x, y, and z in terms of t, and do not rewrite etet as x. dwdtdwdt = Note: You may want to use exp() for the exponential function. Your answer should be an expression in x, y, z, and t; e.g. "3x - 4y" B ) Use part A to evaluate dwdtdwdt when t=0t=0.

Answer 1

Answer:

$\frac{\partial w}{\partial t} = y(e^t) +(x+z)*(cos(t)) - 3y*sin(3t)$

Step-by-step explanation:

First, note that

$\frac{\partial x}{\partial t} = e^{t} \\\frac{\partial y}{\partial t} = cos(t)$

And using the chain rule in one variable

$\frac{\partial z}{\partial t} = -3sin(3t)$

Now remember that the chain rule in several variables sates that

$\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} * \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} * \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} * \frac{\partial z}{\partial t}$

Therefore the chain rule in several variables would look like this.

$\frac{\partial w}{\partial t} = y(e^t) +(x+z)*(cos(t)) - 3y*sin(3t)$

Answer 2

Answer:

At t = 0

dw/dt = x + y + z

Step-by-step explanation:

Since w is a function of x, y, z

dw/dt = w_x • x_t + w_y • y_t + w_z • z_t

Given w = xy + yz

w_x = y

w_y = x + z

w_z = y

Given

x = e^t

y = 2 + sin(t)

z = 2 + cos(3t)

x_t = e^t

y_t = cos(t)

z_t = -3sin(3t)

Using all these,

dw/dt = ye^t + (x + z)cos(t) - 3ysin(t)

At t = 0

dw/dt = y + x + z

dw/dt = x + y + z