Solutions to Uniform Final Exam for Calculus III
This document presents comprehensive solutions to the Calculus III uniform final exam. These solutions are intended to demonstrate the problem-solving process for multivariable calculus, vector calculus, and series analysis.
Section 1: Multivariable Functions and Partial Derivatives
Problem 1: Find all critical points of the function f(x,y) = xy - xy + y and classify them.
Solution:
To find critical points, we first compute the partial derivatives of f(x,y) and set them equal to zero.
Setting both partial derivatives to zero gives us the system of equations:
From the first equation, we get y(2x - y) = 0, so either y = 0 or y = 2x.
Case 1: y = 0
Case 2: y = 2x
When x = 4/3, y = 2(4/3) = 8/3, giving the critical point (4/3, 8/3).
Therefore, our critical points are (0,0) and (4/3, 8/3).
To classify these points, we use the Second Partial Derivative Test. First, let's compute the second partial derivatives:
The discriminant is given by:
At (0,0):
Since D(0,0) = 0, the test is inconclusive at (0,0).
Let's analyze the behavior of the function near (0,0):
- For points on the x-axis (y = 0), f(x,0) = 0 for all x.
- For points on the y-axis (x = 0), f(0,y) = y > 0 for all y 0.
This means the point (0,0) is like a trough along the x-axis but has a minimum along the y-axis, indicating that (0,0) is a saddle point.
At (4/3, 8/3):
fyy(4/3, 8/3) = -2(4/3) + 2 = -2/3
fxy(4/3, 8/3) = 2(4/3) - 2(8/3) = -8/3
D(4/3, 8/3) = (16/3)(-2/3) - (-8/3) = -96/9 = -32/3 < 0
Problem 2: Evaluate the limit lim(x,y)(0,0) (xy)/(x + y) or show that it doesn't exist.
Solution:
We'll assess the limit along different paths to determine if it exists.
Path 1: Along the x-axis (y = 0)
Path 2: Along the y-axis (x = 0)
Path 3: Along the line y = x
Path 4: Along the curve y = x
Section 2: Multiple Integrals
Problem 3: Evaluate the double integral R xy dA over the region R bounded by y = x, y = 2x - x.
Solution:
First, let's find the points of intersection between the two curves:
So, x = 0 or x = 1. The region R is bounded between x = 0 and x = 1.
For each x in [0,1], the y-values range from the lower curve y = x to the upper curve y = 2x - x.
Setting up the double integral:
Evaluating the inner integral with respect to y:
Simplifying:
Now, evaluating the outer integral with respect to x:
Computing the values:
Problem 4: Evaluate the triple integral E z dV, where E is the tetrahedron bounded by x = 0, y = 0, z = 0, and x + y + z = 1.
Solution:
The plane x + y + z = 1 intersects the coordinate axes at (1,0,0), (0,1,0), and (0,0,1), forming a tetrahedron with these vertices and the origin.
For a given x in [0,1], the y-values range from 0 to (1-x). For each (x,y) with y in [0,(1-x)], the z-values range from 0 to (1-x-y).
Setting up the triple integral:
Evaluating the inner integral with respect to z:
Evaluating the middle integral with respect to y:
Evaluating the outer integral with respect to x:
Section 3: Vector Calculus
Problem 5: Evaluate the line integral C F dr, where F(x,y) = (y, x) and C is the curve consisting of the line segment from (0,0) to (1,0) followed by the line segment from (1,0) to (1,1).
Solution:
We'll parameterize each segment of the curve C separately and then add the results.
Segment 1: From (0,0) to (1,0)
dr/dt = (1, 0)
F(r(t)) = F(t, 0) = (0, t) = (0, t)
F(r(t)) dr/dt = (0, t) (1, 0) = 0
01 0 dt = 0
Segment 2: From (1,0) to (1,1)
dr/dt = (0, 1)
F(r(t)) = F(1, t) = (t, 1) = (t, 1)
F(r(t)) dr/dt = (t, 1) (0, 1) = 1
01 1 dt = 1
Adding the results from both segments:
Problem 6: Use Green's Theorem to evaluate C (xy + xy) dx + (x + xy) dy, where C is the boundary of the region between the circles x + y = 1 and x + y = 4 with positive orientation.
Solution:
Green's Theorem states that for a positively oriented, piecewise-smooth, simple closed curve C and the region D bounded by C:
In our case, P = xy + xy and Q = x + xy. Let's compute the partial derivatives:
Therefore, the integrand is:
The region D is the annulus between the circles. It's convenient to use polar coordinates:
- x = r cos()
- y = r sin()
- dA = r dr d
The limits are r from 1 to 2 and from 0 to 2.
Expressing the integrand in polar coordinates:
Evaluating the inner integral with respect to r:
Computing each term separately:
- 12 r cos() dr = [r/3]12 cos() = (8/3 - 1/3)cos() = (7/3)cos()
- 12 r sin() dr = [r/4]12 sin() = (4 - 1/4)sin() = (15/4)sin()
- 12 -2r cos() sin() dr = -2[r/5]12 cos() sin() = -2(32/5 - 1/5)cos() sin() = -(62/5)cos() sin()
Evaluating the outer integral with respect to :
Computing each term separately:
- 02 (7/3)cos() d = (7/3)[sin()]02 = 0
- 02 (15/4)sin() d = (15/4)[/2 - sin(2)/4]02 = (15/4)() = 15/4
- 02 -(62/5)cos() sin() d = -(62/5)[-cos()/3]02 = -(62/5)(0) = 0
Adding up the results:
Section 4: Sequences and Series
Problem 7: Determine whether the series n=1 (n!)/(10n) converges or diverges.
Solution:
We'll use the Ratio Test to determine the convergence of the series.
For our series, an = n!/(10n). Let's compute the limit:
Since the limit is (which is greater than 1), the series diverges by the Ratio Test.
This result makes intuitive sense. The factorial n! grows much faster than 10n, so the terms of the series increase without bound as n increases.
Problem 8: Find the radius of convergence and interval of convergence of the power series n=0 (-1)nxn/(n+1).
Solution:
To find the radius of convergence, we'll use the Ratio Test:
According to the Ratio Test, the series converges when |x| < 1.
Therefore, the radius of convergence is R = 1.
To find the interval of convergence, we need to check the endpoints x = -1 and x = 1 separately.
At x = 1:
This is an alternating series with terms decreasing in magnitude to 0, so it converges by the Alternating Series Test.
At x = -1:
This is the harmonic series, which is known to diverge.
Section 5: Applications of Multiple Integrals
Problem 9: Find the mass of the lamina that occupies the region D bounded by y = x and y = x, if the density function is (x,y) = x + y.
Solution:
Let's find the points of intersection between the curves:
So, x = 0 or x = 1. The region D is bounded between x = 0 and x = 1.
The mass of a lamina with density function (x,y) is given by the double integral:
Evaluating the inner integral with respect to y:
Computing the values:
Now, evaluating the outer integral with respect to x:
Computing the values:
Problem 10: Find the center of mass of the solid tetrahedron bounded by the coordinate planes and the plane x + y + z = 1, assuming constant density ( = 1).
Solution:
For a solid with constant density, the center of mass (x, , z) is given by the formulas:
First, let's calculate the mass M:
Evaluating the inner integral with respect to z:
Evaluating the middle integral with respect to y:
Evaluating the outer integral with respect to x:
So, the mass of the tetrahedron is M = 1/6.
Next, let's calculate the x-coordinate of the center of mass:
Evaluating the inner integral with respect to z:
Evaluating the middle integral with respect to y:
Evaluating the outer integral with respect to x:
Computing the value:
So, E x dV = 0, and x = (1/M) 0 = 0.
By symmetry, we have:
- E y dV = 0, so = (1/M) 0 = 0
- E z dV = 0, so z = (1/M) 0 = 0
Let me carefully recalculate the integral for x:
So, E x dV = 1/24, and x = 6 (1/24) = 1/4.
By the tetrahedron's symmetry, we can deduce that = 1/4 and z = 1/4 as well.
Conclusion: These solutions demonstrate the systematic approach to solving problems in multivariable calculus. Each solution follows a logical progression from understanding the problem to applying appropriate techniques and computing the final answer. Regular practice with these types of problems is essential for mastering Calculus III.
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