Midterm Exam I, Honors Calculus 222
1. Find the anti-derivative
sec4 x dx.
Solution: Using A.34 and A.33, we obtain
1
2
1
2
sec4 x dx = sin x sec3 x +
sec2 dx = sin x sec3 x + tan x + C.
3
3
3
3
2. Us
Honors Calculus II
Homework 10, due May 1, 2013
1. A prism E is bounded by ve planes x = 0, x = 2, y = 0, z = 0,
y + 2z = 1. Find its mass if the mass density is given by f (x, y, z ) = x2 .
2. A cone
Honors Calculus II
Homework 9, due April 24, 2013
1. Evaluate the double integral D f , f (x, y ) = 6x + 2y 2 , and D is the
region bounded by the parabola x = y 2 and the line x = 4.
2. Interchange t
Honors Calculus II
Homework 8
1. Find the absolute minimum and maximum of the function f (x, y ) =
4x 8xy + 2y + 1 on the triangle with vertices (0, 1), (1, 0), (0, 0).
Solution: The critical points a
Honors Calculus II
Homework 7
1. Let F (t) = f (cos t 1, sin t, t). Given that fx (0, 0, 0) = 2, fy (0, 0, 0) = 3,
fz (0, 0, 0) = 1, calculate F (0) using the chain rule.
Solution:
F (t) = fx (cos t1,
Honors Calculus II
Homework 6
1. A function is dened by f (x, y ) = x + y . Find the domain of f , range
of f and draw level curves f (x, y ) = c for c = 1, 2, 3.
Solution: The domain of denition of f
Honors Calculus II
Homework 5
1. Find the point of self-intersection of the parametric curve x = t2 1,
y = t3 4t 1, and nd both tangents at that point.
Solution: We solve the equations
t2 1 = s2 1,
t3
Honors Calculus II
Homework 4
1. Find the equation of the plane containing the points P (2, 4, 3), Q(3, 7, 1),
R(4, 3, 0).
Solution: Introduce the vectors
a =P Q = (1, 3, 4),
b =P R = (2, 1, 3)
and ca
Honors Calculus II
Homework 3
1. Find the angle at the vertex Q of the triangle with vertices at P (1, 3, 5),
Q(1, 1, 2), R(1, 4, 1).
Solution: The angle at Q is the angle between the vectors a =QP =
Honors Calculus II
Homework 2
1
1. Find the area of the region enclosed by the curves y = x2 , y = 1+x2 .
Sketch the region.
Solution: The intersection points are x = a, where a = 51 . The area
2
is
a
Honors Calculus II
Homework 1
1. Use trigonometric substitution to nd the anti-derivative
x3
1 x2 dx.
Solution: We substitute x = sin u and obtain
x3
sin3 u cos u cos u du
1 x2 dx =
sin u(cos2 u cos4