MATH 107 PRACTICE
Please work out each of the given problems.
Credit will be based on the steps that you show towards the final answer.
Show your work.
PROBLEM 1 Please answer the following true or false.
If false, explain why or provide a counter example.
If true, explain why.
A. (11 Points) Suppose that is defined at x = 3 then f '(x) is also defined at x = 3.
False, 3 could be an endpoint of the interval of convergence. The derivative may not preserve the endpoints. An example of this is where
(12 Points) Let x
= x(t), y = y(t)
be parametric equations for a differentiable curve such that x''(-1)
= y''(-1) = 3
, then the curve is concave up at the point
False, the second derivative is not the y''/x'',
False, the second derivative is not the y''/x'', but rather
(12 Points) If f(x,y)
is a differentiable function at the point P, then Dgradf(P)(P)
the directional derivative in the
direction of gradf(P) cannot be negative.
Dgradf(P)(P) = gradf(P) . gradf(P)
which is the square of the magnitude of the gradient, hence positive.
Test the following series for convergence.
If applicable, determine if the series converges absolutely or
A. (17 Points)
We use the limit comparison test with bn = 1/n.
Sbn diverges since it is the harmonic series. Now compute
since the limit is finite, by the limit comparison test the original series diverges also
Use the ratio test to get
By the ratio test, the series converges absolutely.
(35 Points) Determine
the Maclaurin series for the function
PROBLEM 4 (35
Points) Find the interval of
We use the ratio test first
= |5 - 2x| < 1
Now solve the absolute value inequality
5 - 2x = 1 or 5 - 2x = -1
x = 2 or x = 3
Next test the endpoints
For x = 2 the series becomes
Which diverges by the limit test since limit of the terms goes to infinity not zero.
Similarly at x = 3, the series becomes
Again by the limit test the series diverges. We can conclude that the interval of convergence is
PROBLEM 5 (35 Points) Find the length of one of the petals of the graph of the curve
r = 4 sin(3q)
We use the formula
r2 + r'2 = 16sin2(3q) + 144cos2(3q)
Now to find a and b, the petals all begin at the origin. We find
4sin(3q) = 0
3q = 0, p,...
q = 0, p/3,...
Now calculate the integral
PROBLEM 6 (35 Points) Find a unit vector that is perpendicular to the two vectors
3i + 4j - k
and 2i + j + 2k
We find the cross product of the two vectors
= 9i - 8j - 5k
Now find the magnitude
Divide to get
Consider the paraboloid
z = x2 + y2
Adding z2 to both sides, we getz + z2 = x2 + y2 + z2
r cosf + r2 cos2f = r2
Now divide by r to get
cosf + r cos2f = r
Solve for r to get
(35 Points) Let
f(x,y) = x cos(y2)
. Use the chain rule to determine
the angular rate of change of f given that x = r cos q, y = r sin q
fq = fxxq + fyyq
= cos(y2)(-r sin q) - 2xy sin(y2)(r cos q)
= -r cos(r2 sin2q) sin q - 2r3 cos2q sin q sin(r2 sin2q)
PROBLEM 9 (35 Points) Find the equation of the tangent plane to the surface
at the point (2,1,1).
First find the gradient vector which is the normal vector to the plane by taking partial derivatives
grad(F) = <x/2, 2y, -2z> = <1/2,2,-2>
Now use the formula for a plane given a normal vector and a point
<1/2,2,-2> . <x - 2, y - 1, z - 1> = 0
1/2x - 1 + 2y - 2 - 2z + 2 = 0
x - 2 + 4y - 4 - 4z + 2 = 0
x + 4y - 4z = 4
The temperature of a room in your factory can be modeled by the equation f(x,y)
. There is a round table of radius
centered at the origin.
Use the method of Lagrange multipliers to determine the hottest points on
We want to maximize
We want to maximize
f(x,y) = exy
under the constraint
g(x,y) = x2 + y2 - 8 = 0
Lagrange multipliers tells us that
gradf = lgradg
<yexy, xexy> = l<2x, 2y>
This gives us the three equations
yexy = 2lx xexy = 2ly x2 + y2 = 8
Solving the first two equation for l and setting them equal to each other gives
y/x = x/y
x2 = y2
Plugging this back into the third equation gives
y2 + y2 = 8
y = 2
x = 2
f(x,y) = exy
we immediately see that the maximum occurs when the signs are the same, that is the hottest points are at
(2,2) and (-2,-2)