Calculus III Advanced (Course) (13.3) (Homework)
Contents
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.
Internal Links
Parent Article: Calculus III Advanced (Course)