Calculus III 13 Exam 1

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Chapter 13 Exam

From Calculus 10e by Larson and Edwards, p. 960. Exercises 6, 15, 23, 33, 43, 47, 63, 69.

Exercise 6 Sketching a Contour Map

Describe the levels for the function. Sketch a contour map for the surface using level curves for the given \(c\)-values.

\( z=2x^{2}+y^{2}, \: c=1,2,3,4,5 \)

Solution

Contour Map
Figure 1

Level Curves
Figure 2

Exercise 15 Finding Partial Derivatives

Find all first partial derivatives.

\(f(x,y)=5x^{3}+7y-3\)

Solution

\(f_{x}(x,y) = 15x^{2}\)
\(f_{y}(x,y) = 7\)

Exercise 23 Finding Second Partial Derivatives

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

\( f(x,y) = 3x^{2}-xy+2y^{3}\)

Solution

\( f_{x}(x,y) = 6\)
\( f_{y}(x,y) = 12y\)

Exercise 33 Using a Differential as an Approximation

(a) evaluate \(f(2,1)\) and \(f(2.1,1.05)\) and calculate \(\Delta z\).
(b) use the total differential \(dz\) to approximate \(\Delta z\).

\( f(x,y)= 4x+2y\)

Solution

\( (x,y) = (2,1) = (x+\Delta x,y+ \Delta y) = (2.1,1.05)\)

produces

\(dx = \Delta x = 0.1 \text{ and } dy = \Delta y=0.05\)
\( \Delta z \approx dz = 4 \Delta x + 2 \Delta y\)

The approximation for \(f(2,1)\) is

\( \Delta z \approx 4(0.1) + 2 (0.05) = 0.4 + 0.1 = 0.5\)

Exercise 43 Finding a Directional Derivative

Find the directional derivative for the function \(P\) at in the \(v\) direction.

\( f(x,y) = x^{2}y , \: P(-5,5), \: \textbf{v} = 3 \textbf{i} - 4 \textbf{j} \)

Solution
Normal

Exercise 47 Using Gradient Properties

Find the gradient for the function and the maximum value for the directional derivative at the given point.

\(z=x^{2}y \text{, }(2,1) \)

Solution Let \(T(x,y) = x^{2}y \). This produces

\( \nabla T(x,y) \) \(= T_{x}(x,y)\textbf{i} + T_{y}(x,y)\textbf{j}\)
\(= 2xy\textbf{i} + x^{2}\textbf{j}\) Function Gradient
\( \nabla T(2,1) \) \(= 2(2)(1)\textbf{i} + 2^{2}\textbf{j} \)
\(= 4\textbf{i} + 4\textbf{j}\) Direction for maximum increase
\( \nabla T(2,1) \) \(= \sqrt{16 + 16} \)
\(= \sqrt{32}\)

Exercise 63 Using the Second Partials Test

Examine the function for relative extrema and saddle points.

\(f(x,y)=2x^{2}+6xy+9xy^{2}+8x+14\)

Solution

Figure 3

There is no saddle point from inspecting the graph in Figure 3.

\(f_{x}(x,y) = 4x+6y+8 \text{ and } f_{y}(x,y) = 6x+18y\)
\(f_{xx}(x,y) = 4 \text{ and } f_{yy}(x,y) = 18\)
0 \(= 4x+6y+8\)
0 \(= 6x+18y \)
reduces to
\(x\) \(= -\frac{3}{2} \)
\(y\) \(= -\frac{1}{3} \)

Which means the function has on extrema point, negative, at \(f(-1/3,-3/2)\).


Exercise 69 Maximum Revenue

A company manufactures two bicycles types , a racing bicycle and a mountain bicycle. The total revenue from \(x_{1}\) racing bicycle units and \(x_{2}\) mountain bicycle units is

\( R= -6x_{1}^{2} -10x_{2}^{2} -2x_{1}x_{2} + 32x_{1} +84x_{2}\)

where \(x_{1}\) and \(x_{2}\) are in thousands. Find \(x_{1}\) and \(x_{2}\) so that revenue is maximized.
Solution Let \(x_{1} = x\) and \(x_{2} =y\). This rewrites the function as

\( R(x,y) = -6x^{2} -10y^{2} -2xy + 32x +84y\)

This makes the partial derivatives

\( f_{x}(x,y) = 32-12x-2y\)
\( f_{y}(x,y) = 84-2x-20y \)

Setting both equal to zero produces

\(x=2\)
\(y=4\)

The Second Partial Derivatives are:

\( f_{xx}(x,y) = -12\)
\( f_{yy}(x,y) = -20 \)
\( f_{xy}(x,y) = -2 \)

The maximum profit is

\(f(2,4) = -6(4) -10(4) -2(2)(4) + 32(2) +84(4) = 152 \) per bicycle.

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Parent Article: Calculus III Advanced (Course)