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Chapter 12: Q. 34 (page 944)

Find the first-order partial derivatives for the functions in Exercises 27–36.
$g (r, θ) = r \cos θ$

Short Answer

Expert verified

The first-order partial derivatives are $\frac{\partial g}{\partial r} = \cos θ and \frac{\partial g}{\partial θ} = - r \sin θ$

Step by step solution

01

Given

The given function is: $g (r, θ) = r \cos θ$

02

To find

We have to find the first-order partial derivatives.

03

Calculation

$g (r, θ) = r \cos θ \frac{\partial g}{\partial r} = \frac{\partial}{\partial r} (r \cos θ) \frac{\partial g}{\partial r} = \cos θ g (r, θ) = r \cos θ \frac{\partial g}{\partial θ} = \frac{\partial}{\partial θ} (r \cos θ) \frac{\partial g}{\partial θ} = r (- \sin θ) \frac{\partial g}{\partial θ} = - r \sin θ Hence, \frac{\partial g}{\partial r} = \cos θ and \frac{\partial g}{\partial θ} = - r \sin θ$

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Most popular questions from this chapter

$Let z = (f {(x, y))}^{n} . Show that \frac{\partial z}{\partial x} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} and \frac{\partial z}{\partial y} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y} .$

In Exercises $37 - 42$ , use the partial derivatives of role="math" localid="1650186824938" $f (x, y) = x^{y}$ and the point $(e, 3)$ specified to

$(a)$ find the equation of the line tangent to the surface defined by the function in the $x$ direction,

$(b)$ find the equation of the line tangent to the surface defined by the function in the $y$ direction, and

$(c)$ find the equation of the plane containing the lines you found in parts $(a)$ and $(b)$ .

$Extrema : Find the local maxima, local minima, and saddle points of the given functions f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4 .$

In Exercises, by considering the function $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0,$ you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.

Why does the method of Lagrange multipliers fail with this function?

$Let z = f (x, y) g (x, y) . Show that \frac{\partial z}{\partial x} = \frac{f_{x} (x, y) g (x, y) - f (x, y) g_{x} (x, y)}{(g {(x, y))}^{2}} and \frac{\partial z}{\partial y} = \frac{f_{y} (x, y) g (x, y) - f (x, y) g_{y} (x, y)}{(g {(x, y))}^{2}}$

See all solutions

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