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Math 213 Calculus III Practice Exam 2 Solutions
Introduction
Welcome to the solutions page for Math 213 Calculus III Practice Exam 2. This page provides detailed step-by-step solutions for each problem on the practice exam. These solutions are designed to help you understand the concepts and techniques needed to succeed in Calculus III.
Before reviewing the solutions, try attempting each problem on your own first. Then compare your approach and answers with the solutions provided below.
Question 1
Find the limit as (x,y) approaches (0,0) of f(x,y) = (x + y)/(x + y), or show that it does not exist.
Solution:
Step 1: Convert to polar coordinates to simplify the limit.
x = rcos(), y = rsin()
x = rcos(), y = rsin()
Step 2: Substitute into the function:
f(x,y) = (rcos() + rsin())/(rcos() + rsin())
= r(cos() + sin())/r(cos() + sin())
= r(cos() + sin())/(cos() + sin())
f(x,y) = (rcos() + rsin())/(rcos() + rsin())
= r(cos() + sin())/r(cos() + sin())
= r(cos() + sin())/(cos() + sin())
Step 3: Simplify using the identity cos() + sin() = 1:
f(x,y) = r(cos() + sin())
f(x,y) = r(cos() + sin())
Step 4: Evaluate the limit as r approaches 0:
lim[r0] r(cos() + sin()) = 0, since -1 cos() + sin() 1
lim[r0] r(cos() + sin()) = 0, since -1 cos() + sin() 1
Therefore, the limit exists and equals 0.
Question 2
Find and classify all critical points of f(x,y) = x + 3xy + y - 3x.
Solution:
Step 1: Find the partial derivatives:
f(x,y) = 3x + 3y - 3
f(x,y) = 3x + 2y
f(x,y) = 3x + 3y - 3
f(x,y) = 3x + 2y
Step 2: Find critical points by setting partial derivatives to zero:
3x + 3y - 3 = 0 x + y - 1 = 0
3x + 2y = 0 y = -3x/2
3x + 3y - 3 = 0 x + y - 1 = 0
3x + 2y = 0 y = -3x/2
Step 3: Substitute y into the first equation:
x + (-3x/2) - 1 = 0
2x - 3x - 2 = 0
x = (3 (9 + 16))/4 = (3 5)/4
x = 2 or x = -1/2
x + (-3x/2) - 1 = 0
2x - 3x - 2 = 0
x = (3 (9 + 16))/4 = (3 5)/4
x = 2 or x = -1/2
Step 4: Find corresponding y values:
For x = 2: y = -3(2)/2 = -3
For x = -1/2: y = -3(-1/2)/2 = 3/4
For x = 2: y = -3(2)/2 = -3
For x = -1/2: y = -3(-1/2)/2 = 3/4
Step 5: Find the second partial derivatives:
f(x,y) = 6x
f(x,y) = 2
f(x,y) = 3
f(x,y) = 6x
f(x,y) = 2
f(x,y) = 3
Step 6: Use the Hessian determinant to classify critical points:
D = f(x,y)f(x,y) - (f(x,y))
D = f(x,y)f(x,y) - (f(x,y))
At point (2, -3):
D = f(2,-3)f(2,-3) - (f(2,-3)) = 122 - 9 = 24 - 9 = 15 > 0
Since D > 0 and f(2,-3) = 12 > 0, this is a local minimum.
D = f(2,-3)f(2,-3) - (f(2,-3)) = 122 - 9 = 24 - 9 = 15 > 0
Since D > 0 and f(2,-3) = 12 > 0, this is a local minimum.
At point (-1/2, 3/4):
D = f(-1/2,3/4)f(-1/2,3/4) - (f(-1/2,3/4)) = (-3)2 - 9 = -6 - 9 = -15 < 0
Since D < 0, this is a saddle point.
D = f(-1/2,3/4)f(-1/2,3/4) - (f(-1/2,3/4)) = (-3)2 - 9 = -6 - 9 = -15 < 0
Since D < 0, this is a saddle point.
Critical points: (2, -3) is a local minimum, and (-1/2, 3/4) is a saddle point.
Question 3
Evaluate the double integral _R e^(x+y) dA over the region R: x + y 4.
Solution:
Step 1: Identify that the region R is a circle with radius 2 centered at the origin.
Step 2: Convert to polar coordinates:
x + y = r
dA = r dr d
The region becomes: 0 r 2, 0 2
x + y = r
dA = r dr d
The region becomes: 0 r 2, 0 2
Step 3: Rewrite the integral in polar coordinates:
_R e^(x+y) dA = _0^{2} _0^2 e^(r) r dr d
_R e^(x+y) dA = _0^{2} _0^2 e^(r) r dr d
Step 4: Evaluate the inner integral using substitution:
Let u = r, du = 2r dr
_0^2 e^(r) r dr = 1/2 _0^4 e^u du = 1/2 [e^u]_0^4 = 1/2 (e - 1)
Let u = r, du = 2r dr
_0^2 e^(r) r dr = 1/2 _0^4 e^u du = 1/2 [e^u]_0^4 = 1/2 (e - 1)
Step 5: Evaluate the outer integral:
_0^{2} 1/2 (e - 1) d = 1/2 (e - 1) _0^{2} d = 1/2 (e - 1) []_0^{2} = 1/2 (e - 1)2 = (e - 1)
_0^{2} 1/2 (e - 1) d = 1/2 (e - 1) _0^{2} d = 1/2 (e - 1) []_0^{2} = 1/2 (e - 1)2 = (e - 1)
Therefore, _R e^(x+y) dA = (e - 1).
Question 4
Find the directional derivative of f(x,y,z) = xyz at the point (1,2,3) in the direction of v = <1,1,1>.
Solution:
Step 1: Find the gradient of f:
f(x,y,z) = =
f(x,y,z) =
Step 2: Evaluate the gradient at the point (1,2,3):
f(1,2,3) = <23, 13, 12> = <6, 3, 2>
f(1,2,3) = <23, 13, 12> = <6, 3, 2>
Step 3: Find the unit vector in the direction of v:
|v| = (1 + 1 + 1) = 3
u = v/|v| = <1/3, 1/3, 1/3>
|v| = (1 + 1 + 1) = 3
u = v/|v| = <1/3, 1/3, 1/3>
Step 4: Compute the directional derivative:
Df(1,2,3) = f(1,2,3) u
= <6, 3, 2> <1/3, 1/3, 1/3>
= 6/3 + 3/3 + 2/3
= 11/3
= 113/3
Df(1,2,3) = f(1,2,3) u
= <6, 3, 2> <1/3, 1/3, 1/3>
= 6/3 + 3/3 + 2/3
= 11/3
= 113/3
The directional derivative is 113/3.
Question 5
Evaluate _E z dV where E is the tetrahedron bounded by the four planes x=0, y=0, z=0, and x+y+z=1.
Solution:
Step 1: Determine the limits of integration by sketching the tetrahedron:
For the x-y plane: x+y=1
The region in the xy-plane is a triangle with vertices (0,0), (1,0), and (0,1)
For the x-y plane: x+y=1
The region in the xy-plane is a triangle with vertices (0,0), (1,0), and (0,1)
Step 2: Set up the triple integral:
_E z dV = _0^1 _0^{1-x} _0^{1-x-y} z dz dy dx
_E z dV = _0^1 _0^{1-x} _0^{1-x-y} z dz dy dx
Step 3: Evaluate the innermost integral with respect to z:
_0^{1-x-y} z dz = [z/2]_0^{1-x-y} = (1-x-y)/2
_0^{1-x-y} z dz = [z/2]_0^{1-x-y} = (1-x-y)/2
Step 4: Evaluate the integral with respect to y:
_0^{1-x} (1-x-y)/2 dy = 1/2 _0^{1-x} (1-x-y) dy
Let u = 1-x-y, du = -dy
When y = 0, u = 1-x
When y = 1-x, u = 0
= -1/2 _{1-x}^0 u du = 1/2 _0^{1-x} u du
= 1/2 [u/3]_0^{1-x} = (1-x)/6
_0^{1-x} (1-x-y)/2 dy = 1/2 _0^{1-x} (1-x-y) dy
Let u = 1-x-y, du = -dy
When y = 0, u = 1-x
When y = 1-x, u = 0
= -1/2 _{1-x}^0 u du = 1/2 _0^{1-x} u du
= 1/2 [u/3]_0^{1-x} = (1-x)/6
Step 5: Evaluate the integral with respect to x:
_0^1 (1-x)/6 dx = 1/6 _0^1 (1-x) dx
Let u = 1-x, du = -dx
When x = 0, u = 1
When x = 1, u = 0
= -1/6 _1^0 u du = 1/6 _0^1 u du
= 1/6 [u/4]_0^1 = 1/24
_0^1 (1-x)/6 dx = 1/6 _0^1 (1-x) dx
Let u = 1-x, du = -dx
When x = 0, u = 1
When x = 1, u = 0
= -1/6 _1^0 u du = 1/6 _0^1 u du
= 1/6 [u/4]_0^1 = 1/24
Therefore, _E z dV = 1/24.
Question 6
Use Green's Theorem to evaluate _C (y - x)dx + (x + y)dy, where C is the positively oriented circle x + y = 4.
Solution:
Step 1: Recall Green's Theorem:
_C P dx + Q dy = _D (Q/x - P/y) dA
where D is the region enclosed by C
_C P dx + Q dy = _D (Q/x - P/y) dA
where D is the region enclosed by C
Step 2: Identify P and Q:
P(x,y) = y - x
Q(x,y) = x + y
P(x,y) = y - x
Q(x,y) = x + y
Step 3: Find the partial derivatives:
Q/x = /x (x + y) = 2x
P/y = /y (y - x) = 2y
Q/x = /x (x + y) = 2x
P/y = /y (y - x) = 2y
Step 4: Apply Green's Theorem:
_C (y - x)dx + (x + y)dy = _D (2x - 2y) dA
_C (y - x)dx + (x + y)dy = _D (2x - 2y) dA
Step 5: Convert to polar coordinates:
The region D is a circle of radius 2.
x = rcos(), y = rsin(), dA = r dr d
_D (2x - 2y) dA = _0^{2} _0^2 (2rcos() - 2rsin()) r dr d
The region D is a circle of radius 2.
x = rcos(), y = rsin(), dA = r dr d
_D (2x - 2y) dA = _0^{2} _0^2 (2rcos() - 2rsin()) r dr d
Step 6: Evaluate the integral:
= _0^{2} _0^2 (2rcos() - 2rsin()) dr d
= 2 _0^{2} _0^2 (rcos() - rsin()) dr d
= 2 _0^{2} (cos() - sin()) [r/3]_0^2 d
= 2 _0^{2} (cos() - sin())(8/3) d
= (16/3) _0^{2} (cos() - sin()) d
= (16/3) [sin() + cos()]_0^{2}
= (16/3) [(sin(2) + cos(2)) - (sin(0) + cos(0))]
= (16/3) [(0 + 1) - (0 + 1)]
= (16/3)0 = 0
= _0^{2} _0^2 (2rcos() - 2rsin()) dr d
= 2 _0^{2} _0^2 (rcos() - rsin()) dr d
= 2 _0^{2} (cos() - sin()) [r/3]_0^2 d
= 2 _0^{2} (cos() - sin())(8/3) d
= (16/3) _0^{2} (cos() - sin()) d
= (16/3) [sin() + cos()]_0^{2}
= (16/3) [(sin(2) + cos(2)) - (sin(0) + cos(0))]
= (16/3) [(0 + 1) - (0 + 1)]
= (16/3)0 = 0
Therefore, _C (y - x)dx + (x + y)dy = 0.
Tips for Calculus III Exams
- Understand fundamental concepts: Make sure you have a solid grasp of limits, partial derivatives, and multiple integrals before attempting more complex problems.
- Practice visualization: Developing the ability to visualize 3D shapes and functions will greatly aid your understanding of flux, divergence, and surface integrals.
- Master coordinate transformations: Polar, cylindrical, and spherical coordinates often simplify multivariable problems significantly.
- Know your theorems: Green's, Stokes', and the Divergence Theorems provide powerful tools for evaluating complex line and surface integrals.
- Work through examples: The best way to prepare is to work through a variety of practice problems, ensuring you understand each step.
- Check your work: Always plug your answers back into the original equations or verify that they satisfy the given conditions when possible.
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