Admin
13 Jun 2026 04:52
admin
#general
Math253 Calculus III Winter 2019 Practice Final Solutions
These solutions cover the practice final examination for Math253 Calculus III from Winter 2019. The solutions provide step-by-step explanations to help you understand the concepts and techniques needed to solve similar problems.
Problem 1: Partial Derivatives
Find all first and second partial derivatives of the function f(x,y,z) = excos(y) + ln(z).
First partial derivatives:
f/x = excos(y)
f/y = -exsin(y)
f/z = 1/z
Second partial derivatives:
f/x = excos(y)
f/y = -excos(y)
f/z = -1/z
f/xy = f/yx = -exsin(y)
f/xz = f/zx = 0
f/yz = f/zy = 0
Problem 2: Multiple Integrals
Evaluate the double integral R (x + y) dA, where R is the region bounded by x = y and x - 2y = 3.
First, we need to find the intersection points of the curves:
y = x and x - 2y = 3
y = 2y + 3 y - 2y - 3 = 0 (y - 3)(y + 1) = 0
The intersections occur at y = -1 and y = 3
Setting up the integral with respect to y first:
-13 y2y+3 (x + y) dx dy
Evaluating the inner integral:
y2y+3 (x + y) dx = [x + xy]y2y+3
= (2y+3) + (2y+3)y - (y) - y(y)
= (4y + 12y + 9) + 2y + 3y - y - y
= 2y + 6y + 4.5 + 2y + 3y - y - y
= -y - y + 4y + 9y + 4.5
Now evaluating the outer integral:
-13 (-y - y + 4y + 9y + 4.5) dy
= [-y - y + y + y + 4.5y]-13
= [-(3) - (3) + (3) + (3) + 4.5(3)] - [-(-1) - (-1) + (-1) + (-1) + 4.5(-1)]
= [-(243) - (81) + (27) + (9) + 13.5] + [ - - + - 4.5]
= [-24.3 - 20.25 + 36 + 40.5 + 13.5] + [0.1 - 0.25 - 1.33 + 4.5 - 4.5]
= 45.45 + (-1.48) = 43.97
Problem 3: Vector Calculus
Find the unit tangent vector and the principal unit normal vector for the curve r(t) = (3cos(t), 3sin(t), 4t) at t = /2.
First, find the derivative of r(t):
r'(t) = (-3sin(t), 3cos(t), 4)
Find the magnitude of r'(t):
|r'(t)| = [9sin(t) + 9cos(t) + 16] = [9(sin(t) + cos(t)) + 16] = [9 + 16] = 5
Find the unit tangent vector T(t):
T(t) = r'(t)/|r'(t)| = (-3sin(t)/5, 3cos(t)/5, 4/5)
Evaluate T(t) at t = /2:
T(/2) = (-3sin(/2)/5, 3cos(/2)/5, 4/5) = (-3/5, 0, 4/5)
Find the derivative of T(t):
T'(t) = (-3cos(t)/5, -3sin(t)/5, 0)
Find the magnitude of T'(t):
|T'(t)| = [9cos(t)/25 + 9sin(t)/25] = [9(cos(t) + sin(t))/25] = 3/5
Find the principal unit normal vector N(t):
N(t) = T'(t)/|T'(t)| = (-3cos(t)/5)/(3/5), -3sin(t)/5)/(3/5), 0) = (-cos(t), -sin(t), 0)
Evaluate N(t) at t = /2:
N(/2) = (-cos(/2), -sin(/2), 0) = (0, -1, 0)
Problem 4: Line and Surface Integrals
Evaluate the line integral C F dr, where F = (y, -x, 0) and C is the circle x + y = 4, z = 0, oriented counterclockwise when viewed from above.
We can parameterize the circle C:
r(t) = (2cos(t), 2sin(t), 0) for 0 t 2
Find the derivative of r(t):
r'(t) = (-2sin(t), 2cos(t), 0)
Express F in terms of the parameterization:
F(r(t)) = (2sin(t), -2cos(t), 0)
Compute the dot product F r':
F(r(t)) r'(t) = (2sin(t))(-2sin(t)) + (-2cos(t))(2cos(t)) + 0
= -4sin(t) - 4cos(t) = -4(sin(t) + cos(t)) = -4
Evaluate the line integral:
C F dr = 02 (-4) dt = -4t|02 = -8
Alternatively, we could use Green's theorem:
C F dr = D (Q/x - P/y) dA
= D (-1 - 1) dA = -2 D dA = -2 (area of circle with radius 2) = -2 4 = -8
Problem 5: Surface Integrals
Evaluate S F dS, where F = (x, y, z) and S is the surface of the sphere x + y + z = 4, oriented outward.
We can use the divergence theorem:
S F dS = V ( F) dV
Calculate the divergence of F:
F = /x(x) + /y(y) + /z(z) = 1 + 1 + 1 = 3
Set up the triple integral in spherical coordinates:
V 3 dV = 02 0 02 3 sin() d d d
Evaluate the integral:
= 3 02 0 02 sin() d d d
= 3 02 0 [sin()]02 d d
= 3 02 0 sin() d d
= 8 02 [-cos()]0 d
= 8 02 (-cos() + cos(0)) d
= 8 02 2 d
= 16 02 d = 16|02 = 32
Problem 6: Optimization
Find the maximum and minimum values of the function f(x,y) = x + y + 1 on the closed disk D = {(x,y) | x + y 4}.
First, find critical points in the interior of D:
f/x = 2x = 0 x = 0
f/y = 2y = 0 y = 0
The only critical point in the interior is (0,0) with f(0,0) = 1
Next, find extreme values on the boundary x + y = 4:
We can use Lagrange multipliers. Let g(x,y) = x + y = 4.
f = g gives (2x, 2y) = (2x, 2y)
This gives x(1 - ) = 0 and y(1 - ) = 0
This yields two possibilities:
- If = 1, then any point on the circle x + y = 4 is a candidate.
- If x = 0, then y = 4 y = 2. If y = 0, then x = 4 x = 2.
Evaluate f at these points:
f(2, 0) = 4 + 0 + 1 = 5
f(0, 2) = 0 + 4 + 1 = 5
For any point on the circle x + y = 4, f(x,y) = x + y + 1 = 4 + 1 = 5
Comparing values:
f(0,0) = 1 (minimum)
f(x,y) = 5 for any point on the boundary (maximum)
Therefore:
The minimum value of f is 1 at (0,0)
The maximum value of f is 5 at all points on the circle x + y = 4
Key Concepts Review
For the Math253 Calculus III final exam, ensure you have mastered these key concepts:
- Partial Derivatives: Understanding how to calculate first and second partial derivatives and interpret their meaning.
- Multiple Integrals: Evaluating double and triple integrals using appropriate coordinate systems and determining proper limits of integration.
- Vector Calculus: Computing gradient, divergence, and curl, and understanding their physical interpretations.
- Line Integrals: Evaluating line integrals of scalar and vector fields, and understanding their relationships to work and flow.
- Surface Integrals: Calculating flux of vector fields through surfaces and using the divergence theorem.
- Green's, Stokes', and Divergence Theorems: Applying these fundamental theorems to relate different types of integrals.
- Optimization in Multiple Variables: Finding critical points and applying Lagrange multipliers for constrained optimization.
Practice working through problems similar to those in this practice final, focusing on understanding the underlying concepts rather than just memorizing procedures. Remember to check your work and verify that your answers make sense in the context of the problem.
Reference Files For Math253 Calculus III Winter 2019 Practice Final Solutions
Screenshoot
Description
This file is just a reference file for Math253 Calculus III Winter 2019 Practice Final Solutions. Does not guarantee that the specific things you want are included in it.
Direct download (wait 10 seconds)
Math253 Calculus III Winter 2019 Practice Final Solutions and Reference File Download Link
Admin
2026-06-13 04:52:16
MATH253 Calculus III Master Syllabus 2019-2020 and Reference File Download Link
Admin
2026-06-08 09:20:17
Calculus III Practice Final Exam Solutions and Reference File Download Link
Admin
2026-06-12 21:28:15
Solutions To Uniform Final Exam For Calculus III and Reference File Download Link
Admin
2026-06-13 02:16:15
Cookie Consent
We use cookies to enhance your browsing experience and analyze site traffic. By clicking 'Accept all cookies', you agree to the use of these cookies. You can manage your preferences or learn more in our [Privacy Policy/Cookie Policy.