Please help with D. The function is in the first photo 5-4f3.2+412 35 6Figure 10.6.1 A contour plot of f(x, y) = 30-1²-². CHAPTER 10. DERIVATIVES OF MULTIVARIABLE FUNCTIONSc. Now, rather than walking due east or due north, let's suppose that theperson is walking with velocity given by the vector v = (3,4), where timeis measured in seconds. Note that the person's speed is thus |v|= 5 feetper second. Find parametric equations for the person's path; that is,parametrize the line through (2, 1) using the direction vector v= (3, 4).Let r(t) denote the z-coordinate of the line, and y(t) its y-coordinate.Make sure your parametrization places the walker at the point (2, 1) whent=0.+ = 2² (+₁4) =(²11) Whent=02 = ²+² +²²³(ky) = (2₂1) + (3,4)(x,y) = (2, 1) + (3=14€)(x47) = (2+ 3+₁ 1+4)x=2+3ty=1+4681EQN=X(t) = 2+ 3ty (=/+4td. With the parametrization in (c), we can now view the temperature fas not only a function of z and y, but also of time, t. Hence, use thechain rule to determine the value of 0 What are the units on youranswer? What is the practical meaning of this result?

A graph of contour plot is given as f(x, y)=30-x2-12y2. The parametric equations obtained in part…

Answered: d. With the parametrization in (c), we…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Please help with D. The function is in the first photo

5-
4f
3.
2+
4
1
2 3
5 6
Figure 10.6.1 A contour plot of f(x, y) = 30-1²-².

CHAPTER 10. DERIVATIVES OF MULTIVARIABLE FUNCTIONS
c. Now, rather than walking due east or due north, let's suppose that the
person is walking with velocity given by the vector v = (3,4), where time
is measured in seconds. Note that the person's speed is thus |v|= 5 feet
per second. Find parametric equations for the person's path; that is,
parametrize the line through (2, 1) using the direction vector v= (3, 4).
Let r(t) denote the z-coordinate of the line, and y(t) its y-coordinate.
Make sure your parametrization places the walker at the point (2, 1) when
t=0.
+ = 2² (+₁4) =(²11) Whent=0
2 = ²+² +²²³
(ky) = (2₂1) + (3,4)
(x,y) = (2, 1) + (3=14€)
(x47) = (2+ 3+₁ 1+4)
x=2+3t
y=1+46
81
EQN=
X(t) = 2+ 3ty (=/+4t
d. With the parametrization in (c), we can now view the temperature f
as not only a function of z and y, but also of time, t. Hence, use the
chain rule to determine the value of 0 What are the units on your
answer? What is the practical meaning of this result?

Expert Solution

Step 1

A graph of contour plot is given as $f (x, y) = 30 - x^{2} - \frac{1}{2} y^{2}$ .

The parametric equations obtained in part (c) as

$x (t) = 2 + 3 t y (t) = 1 + 4 t$

The aim is to find the derivative ${\frac{d f}{d t}}_{t = 0}$ using chain rule.

Let $x (t)$ and $y (t)$ are differentiable functions in $t$ , $f (x, y)$ is a differentiable function in $x and y$ . Then $f (x (t), y (t))$ is a differentiable function in $t$ and the chain rule of derivative states that $\frac{d f}{d t} = \frac{\partial f}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial f}{\partial y} \cdot \frac{d y}{d t}$ .