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Chapter 13: Problem 19

Find the following derivatives. $$z_{s} \text { and } z_{t}, \text { where } z=x^{2} \sin y, x=s-t, \text { and } y=t^{2}$$

Short Answer

Expert verified
Question: Find the partial derivatives of the function $$z = x^2 \sin y$$ with respect to s and t, given $$x = s - t$$ and $$y = t^2$$. Answer: The partial derivatives of z with respect to s and t are $$z_s = 2x \sin y$$ and $$z_t = -2x \sin y + 2tx^2 \cos y$$.

Step by step solution

01

Write down the given function and chain rule for partial derivatives.

The given function is: $$z = x^2 \sin y,$$ where $$x = s - t$$ and $$y = t^2.$$ We are asked to find $$z_s$$ and $$z_t$$, which represent the partial derivatives of z with respect to s and t. The chain rule for partial derivatives states that: $$z_s = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s},$$ and $$z_t = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}.$$
02

Find the partial derivatives of z with respect to x and y.

To find $$z_s$$ and $$z_t$$, we need first to compute the partial derivatives of z. Based on the given function, we have: $$\frac{\partial z}{\partial x} = 2x \sin y,$$ and $$\frac{\partial z}{\partial y} = x^2 \cos y.$$
03

Find the partial derivatives of x and y with respect to s and t.

Based on the given functions for x and y, we have: $$\frac{\partial x}{\partial s} = 1,$$ $$\frac{\partial x}{\partial t} = -1,$$ and $$\frac{\partial y}{\partial s} = 0,$$ $$\frac{\partial y}{\partial t} = 2t.$$
04

Find z_s using the chain rule.

Now, substitute the partial derivatives found in steps 2 and 3 into the formula for $$z_s$$: $$z_s = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}.$$ $$z_s = (2x \sin y)(1) + (x^2 \cos y)(0).$$ Removing the terms with a zero in them, we get: $$z_s = 2x \sin y.$$
05

Find z_t using the chain rule.

Now, substitute the partial derivatives found in steps 2 and 3 into the formula for $$z_t$$: $$z_t = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}.$$ $$z_t = (2x \sin y)(-1) + (x^2 \cos y)(2t).$$ Simplifying, we get: $$z_t = -2x \sin y + 2tx^2 \cos y.$$ The final answer for the partial derivatives of z with respect to s and t are: $$z_s = 2x \sin y,$$ and $$z_t = -2x \sin y + 2tx^2 \cos y.$$

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