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Chapter 12: Q 4. (page 963)

Let $z = e 鈭� x y 2, x = s \sin t,$ and $y = s 2 \cos t$
(a) Find $\frac{鈭� z}{鈭� s}$
by using the Chain Rule, Theorem $12.33$
(b) Find $\frac{鈭� z}{鈭� s}$ by evaluating $f (x (s, t), y (s, t)) = f (s \sin t, s 2 \cos t)$ and taking the partial derivative with respect to $s$ of the resulting function.
(c) Show that your answers from parts (a) and (b) are the same. Which method was easier?

Short Answer

Expert verified

Part (a) $- e^{- (s^{3} \sin^{2} t \cos^{2} t)} (5 s^{4} \sin t \cos^{2} t)$

Part (b) $- e^{- (s^{5} \sin t \cos^{2} t)} (5 s^{4} \sin t \cos^{2} t)$

Part (c) The method based on the chain rule is less difficult than the method used in part (b).

Step by step solution

01

Part (a) Step 1: Given information

$z = e^{- x y^{2}}, x = s \sin t$ and $y = s^{2} \cos t$

02

Part (a) Step 1: Explanation

Let $z = e^{- x y^{2}}, x = s \sin t$ and $y = s^{2} \cos t$

The goal is to locate $\frac{鈭� z}{鈭� s}$ using chain rule.

According to the chain rule

$\frac{鈭� z}{鈭� s} = \frac{鈭� z}{鈭� x} \frac{鈭� x}{鈭� s} + \frac{鈭� z}{鈭� y} \frac{鈭� y}{鈭� s}$

where $z = f (x, y), x = u (s, t)$ and $y = v (s, t)$

First, find $\frac{鈭� z}{鈭� x}$

$\frac{鈭� z}{鈭� x} = \frac{鈭�}{鈭� x} (e^{- x y^{2}}) = e^{- x y^{2}} (- y^{2}) \frac{鈭�}{鈭� x} x = - y^{2} e^{- x y^{2}}$

Next find $\frac{鈭� z}{鈭� y}$

$\frac{鈭� z}{鈭� y} = \frac{鈭�}{鈭� y} (e^{- x y^{2}}) = e^{- x y^{2}} (- x) \frac{鈭�}{鈭� y} y^{2} = - 2 x y e^{- x y^{2}}$

Again, find $\frac{鈭� x}{鈭� s}$

$\frac{鈭� x}{鈭� s} = \frac{鈭�}{鈭� s} (s \sin t) = \sin t \frac{鈭�}{鈭� s} s = \sin t$

Also, find $\frac{鈭� y}{鈭� s}$

\frac{鈭� y}{鈭� s} = \frac{鈭�}{鈭� s} (s^{2} \cos t) = \cos t \frac{鈭�}{鈭� s} s^{2} = 2 s \cos t

Thus,

\frac{鈭� z}{鈭� s} = \frac{鈭� z}{鈭� x} \frac{鈭� x}{鈭� s} + \frac{鈭� z}{鈭� y} \frac{鈭� y}{鈭� s} = (- y^{2} e^{- x y^{2}}) \sin t + (- 2 x y e^{- x y^{2}}) (2 s \cos t) = - e^{- x y^{2}} (y^{2} \sin t + 4 x y s \cos t) = - e^{- x \sin t {(x^{2} \cos t)}^{2}} \{{(s^{2} \cos t)}^{2} \sin t + 4 (s \sin t) (s^{2} \cos t) s \cos t\} = - e^{- s \sin ((s^{4} {con}^{2} t)} (s^{4} \cos^{2} t \sin t + 4 s^{4} \sin t \cos^{2} t) = - e^{- (s^{3} \sin^{2} t \cos^{2} t)} (5 s^{4} \sin t \cos^{2} t)

03

Part (b) Step 1: Explanation

The goal is to find $\frac{鈭� z}{鈭� s}$

Rewrite the function as

z = e^{- x y^{2}} = e^{- s \sin {(s^{2} \cos^{2})}^{2}} = e^{- s \sin (s^{2} \cos^{2} t)} = e^{- (x^{3} \sin (\cos^{2} t)} 鈥� 鈥� . (1)

Differentiate ( 1 ) partially with respect to $s$

\frac{鈭� z}{鈭� s} = \frac{鈭�}{鈭� s} e^{- (s^{3} \sin^{2} \cos^{2} t)} = e^{- (s^{3} \sin^{2} f \cos^{2} t)} \frac{鈭�}{鈭� s} \{- (s^{5} \sin t \cos^{2} t)\} = e^{- (s^{5} \sin t \cos^{2} t)} (- \sin t \cos^{2} t) \frac{鈭�}{鈭� s} s^{5} = - e^{- (s^{5} \sin^{2} t \cos^{2} t)} (5 s^{4} \sin t \cos^{2} t)

04

Part (c) Step 1: Explanation

Parts (a) and (b) show that the outcome is the same. The method based on the chain rule is less difficult than the method used in part (b).

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Most popular questions from this chapter

$A chain rule for functions of two variables : Let f (x, y) = x^{2} y^{3}, x = s \cos t, and y = s \sin t . Find \frac{鈭� f}{鈭� x} and \frac{鈭� f}{鈭� y} .$

Explain how you could use the method of Lagrange multipliers to find the extrema of a function of two variables, $f (x, y),$ subject to the constraint that $(x, y)$ is a point on the boundary of a triangle $T$ in the xy-plane.

Prove that if you minimize the square of the distance from the origin to a point (x, y) subject to the constraint $g (x, y) = 0$ , you have minimized the distance from the origin to (x, y) subject to the same constraint.

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable. $\frac{d x}{d t} w h e n x = r \cos 胃, r = t^{2} 鈭� 5, a n d 胃 = t^{3} + 1$

In Exercises 21鈥�26, find the discriminant of the given function.

$g (x, y) = e^{- 3 x} \sin (2 y)$ .

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