Calculus III 13 Exam 1
Contents
- 1 Chapter 13 Exam
- 1.1 Exercise 6 Sketching a Contour Map
- 1.2 Exercise 15 Finding Partial Derivatives
- 1.3 Exercise 23 Finding Second Partial Derivatives
- 1.4 Exercise 33 Using a Differential as an Approximation
- 1.5 Exercise 43 Finding a Directional Derivative
- 1.6 Exercise 47 Using Gradient Properties
- 1.7 Exercise 63 Using the Second Partials Test
- 1.8 Exercise 69 Maximum Revenue
- 2 Internal Links
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
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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
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There is no saddle point from inspecting the graph in Figure 3.
Which means the function has on extrema point, negative, at \(f(-1/3,-3/2)\).
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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.
Internal Links
Parent Article: Calculus III Advanced (Course)