Calculus III Advanced (Course) (13.3) (Homework)

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Section 13.3 Homework

From Calculus 10e by Larson and Edwards, p. 896. Exercises 18, 52, 60, 66, 78, 110.

Exercise 13.3.18 Finding Partial Derivatives

Find both first partial derivatives.

\( z=ye^{y/x} \)

Solution[1] a.

\( f(x,y) = ye^{y/x} \)

The partial derivative with respect to \(x\) is

$$ \frac{\partial }{\partial x} ye^{y/x} = - \frac{1}{x^{2}} e^{y/x}y^{2} $$

The partial derivative with respect to \(y\) is

$$ \frac{\partial }{\partial y} ye^{y/x} = e^{y/x} + \frac{1}{x} ye^{y/x} $$

Exercise 13.3.52 Finding the Slopes of a Surface

find the slopes of the surface in the \(x\)- and \(y\)-directions at the given point.

\( h(x,y) = x^{2} - y^{2} \: (-2,1,3) \)

Solution

\( f_{x}(x,y) = 2x \)

and

\( f_{y}(x,y) = -2y \)
\( f_{x}(-2,1,3) = 4 \)
\( f_{y}(-2,1,3) = -2 \)

Exercise 13.3.60 Evaluating Partial Derivatives

Evaluate \(f_{x}\), \(f_{y}\), \(f_{z}\) at the given point.

\( f(x,y,z) = x^{2}y^{3}+2xyz - 3yz, \: (-2,1,2) \)

Solution

\( f_{x}(x,y,z) = 2xy^{3}+ 2yz\)[2]
\( f_{x}(-2,1,2) = -4y^{3}+2yz\)
\( f_{y}(x,y,z) = 3x^{2}y^{2}+2xz-3z \)
\( f_{y}(-2,1,2) = -4+2z\)
\( f_{z}(x,y,z) = 2xy-3y \)
\( f_{z}(-2,1,2) = \) the point does not exist.

Exercise 13.3.66 Using First Partial Derivatives

For \(f(x,y)\), find all values for \(x\) and \(y\) such that \(f_{x}(x,y) = 0\) and \(f_{y}(x,y) = 0\) simultaneously.

\( f(x,y) = x^{2}-xy + y^{2} -5x + y \)

Solution

\( f_{x}(x,y) = 2x -y-5 \)
\( f_{y}(x,y) = -x+2y+1\)
$$ \left\{\begin{matrix} 2x -y-5 = 0 & \\ -x+2y+1 = 0 & \end{matrix}\right. $$ \(= \Rightarrow x=3, \: y=1 \)

Exercise 13.3.78 Finding Second Partial Derivatives

Find the four second partial derivatives. Observe the second mixed partials are equal.

\( z= \ln(x-y) \)

Solution

$$ \frac{d^{2}}{dx^{2}} = - \frac{1}{(x-y)^{2}} $$
$$ \frac{d^{2}}{dy^{2}} = - \frac{1}{(x-y)^{2}} $$

Exercise 13.3.110 Marginal Costs

\( C= 32 \sqrt{xy} + 175x + 205y + 1050 \)

a. Find the marginal costs when \(x=80\) and \(y=20\).
b. Which increases at a higher rate? How can this be determined from the cost model?
Solution a.

$$ f_{x} = \frac{16\sqrt{y}}{\sqrt{x}} +175 $$
$$ \frac{16\sqrt{20}}{\sqrt{80}} +175 $$
$$ f_{y} = \frac{16\sqrt{y}}{\sqrt{x}} +205 $$
$$ \frac{16\sqrt{20}}{\sqrt{80}} +205 $$

b. The variable \(y\) increases at the higher rate because each unit adds 30 to the cost.

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Parent Article: Calculus III Advanced (Course)