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Chapter 12: TF. 3 (page 946)

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

Short Answer

Expert verified

$\frac{\partial y}{\partial s} = \sin t \frac{\partial y}{\partial t} = s \cos t$

Step by step solution

Step 1. Given information

$f (x, y) = x^{2} y^{3}, x = s \cos t, and y = s \sin t .$

Step 2. Finding partial derivatives

$y = s \sin t . \Rightarrow \frac{\partial y}{\partial s} = \sin t \Rightarrow \frac{\partial y}{\partial t} = s \cos t$

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Most popular questions from this chapter

Fill in the blanks to complete the limit rules. You may assume that $lim_{x \to a} f (x)$ and $lim_{x \to a} g (x)$ exists and that k is a scalar.

$lim_{x \to a} (f (x) - g (x)) =$

In Exercises $37 - 42$ , use the partial derivatives of role="math" localid="1650186853142" $g (x, y) = \tan^{- 1} (x y^{2})$ and the point role="math" localid="1650186870407" $(1, 0)$ 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 $(b)$ .

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.

Explain why $(0, 0)$ is not an extremum of $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0 .$

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 $(b)$ .

Solve the exact differential equations in Exercises 63–66. $e^{x} \ln y + x^{3} + \frac{e^{x}}{y} \frac{d y}{d x} = 0$

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Achainruleforfunctionsoftwovariables:Letf(x,y)=x2y3,x=scost,andy=ssint.Find∂y∂sand∂y∂t.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding partial derivatives

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