Name
MATH 107 PRACTICE
FINAL 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
False, 3 could be an endpoint of the interval of convergence. The derivative may not preserve the endpoints. An example of this is where
1
B.
(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
(x(-1),y(-1))
. Solution
d2y
(y'/x')'
C.
(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. True, since Dgradf(P)(P) = gradf(P) . gradf(P) which is the square of the magnitude of the gradient, hence positive.
PROBLEM 2
Test the following series for convergence.
If applicable, determine if the series converges absolutely or
conditionally. 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
B.
(18 Points)
Use the ratio test to get
By the ratio test, the series converges absolutely.
PROBLEM 3
(35 Points) Determine
the Maclaurin series for the function
3 We write
3
3 1
PROBLEM 4 (35
Points) Find the interval of
convergence 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 (2,3)
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
We have 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
= 8.91
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
PROBLEM 7
(35 Points)
Consider the paraboloid
z = x2 + y2
Adding z2 to both sides, we get z + z2 = x2 + y2 + z2 r cosf + r2 cos2f = r2 Now divide by r to get cosf + r cos2f = r Solve for r to get
cosf
PROBLEM 8
(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
. We want fq(r,q) We have 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
x2 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 x - 2 + 4y - 4 - 4z + 2 = 0 x + 4y - 4z = 4
PROBLEM 10
(35 Points)
The temperature of a room in your factory can be modeled by the equation f(x,y)
= exy
. There is a round table of radius 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
yexy
xexy or y/x = x/y so that x2 = y2 Plugging this back into the third equation gives y2 + y2 = 8 or
y = and
x = From 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) |