Solutions to Sample Final Exam Problems MAT 131, Fall 1998
As always, these solutions to the sample may contain typos.
They are as correct as I could make them when I wrote them, but errors
always creep in. However, the method of solution is certainly correct, so
any errors will be only of a typographical or algebraic nature, which you
should easily be able to pick out if you have already attempted the
problem. And if you haven't attempted the problem, you shouldn't be reading
these solutions anyway!
1. Write the equation of the linear function f with f(0)=1 and
f(3)=3. Also write the equation of the exponential function g with
g(0)=1 and g(3)=3.
Solution:
The linear function must be of the form
f(x) = ax+b. If f(0)=1,
then b=1. The fact that f(3)=3 gives us 3a + 1 =3, so a=2/3.
Thus, the linear equation is
f(x) = 2x/3 + 1.
An exponential function is of the form
g(x)= ke^{cx}. Since
g(0)=1, we have 1 = ke^{0}, so k=1. Now using g(3) = 3 gives
us
. Taking the logarithm of both sides yields
. Thus
, giving
. Notice that this reduces to
g(x)=3^{x/3}.
2.
Compute the derivatives with respect to x for each of the following:
 (a)

x^{1/2} + x + x^{1/2}
 (b)

 (c)

 (c)

 (d)

 (e)

 (f)

 (g)

 (h)

Solution:
 (a)

 (b)

.
 (c)
 Since
,
the derivative is 0.
 (d)

 (e)
 To do this, we use the Fundamental Theorem of
Calculus.
.
 (f)

.
 (g)

.
 (h)

.
3.
Compute each of the following antiderivatives (indefinite integrals):
 (a)

 (b)

 (c)

 (d)

 (e)

 (f)

Solution:
 (a)

.
 (b)

.
 (c)
 To compute
,
first we notice that
because of the ,
we must have something involving a .
Since
,
we are only off by a
constant factor. Thus, the antiderivative must be
.
 (d)

requires a bit of
thought. Rewriting it as
makes the answer more
apparent, however. If we guess that the antiderivative is
and check by taking the derivative, we see it works
perfectly.
 (e)

.
If we hadn't just seen this derivative in problem 1(f), this would be
harder. But we did, so the answer is obviously
.
 (f)
 To compute
,
we guess that
something like e^{2x+1} would work. Taking the derivative of that
shows that it is off by a factor of 2, so the correct antiderivative
must be
4.
Evaluate each of the following definite integrals:
 (a)

 (b)

 (c)

 (d)

 (e)

 (f)

Solution:
 (a)

.
 (b)

.
 (c)

.
 (d)
 (Unfortunately, there was a typo on this problem, so its a
bit uglier than expected. Oh well, we'll do it anyway.)
To calculate
,
first notice that the derivative of the bottom is 3x^{2}, so
is an antiderivative of
.
Thus, the value of the integral is
.
(Sometimes typos just happen, and we have to deal with them. Let's
hope there aren't any on the real final.)
 (e)
 The integral
looks
really hard, unless you realize what it is the area of. This is the
area of the upper half of a circle of radius 1, so its value must be
.
 (f)
 To compute
,
we can either
realize this as the area of two triangles of base 2 and height 2or we can write it as
.
In either case, we get 2+2 = 4 as the result.
5. What is the average value of y=x^{3} over the interval [0,2]?
Solution:
The average value is given by
6. Find the point on the graph of the curve
that is
closest to the point (5,0). (Hint: if d is the distance from
(5,0) to a point on the curve, then it is permissible (and easier!)
to minimize d^{2}.)
Solution:
Let D(x) be the square of the distance from (5,0) to a point
on the graph of
. Then
To find the closest point, we find the minimum of D(x), which must
occur at either x=0 or a critical point of D.
D'(x) = 2x  11,
so the only critical point is x=11/2. This is clearly a minimum,
since D(x) is a parabola, opening upwards. The closest point occurs at
.
7. Give the lefthand sum, righthand sum, and trapezoid
approximation for the integral
,
using n=4 rectangles. What should n be to ensure that the
righthand sum is accurate to within 0.001? (Hint: compare the
expressions for the lefthand and righthand sums for arbitrary n
what does this tell you about the exact value of the integral?)
Solution:
If we are using 4 rectangles, our points are x_{0}=0, x_{1}=1/2,
x_{2}=1, x_{3}=3/2, and x_{4}=2. The lefthand sum is given by
The righthand sum is
and the trapezoid rule is the average of the two, that is,
the trapezoid rule gives
To determine what n should be to ensure that the leftsum has an
error of no more than 0.001, first notice that since
is an increasing function, we always have
This means that the error in either one can be at most R_{n}  L_{n},
and is probably even less than that. So we must find n so that
because all the middle terms cancel out. This means that if we take
, we will be
guaranteed to have an answer that is within 0.001 of the value of
the integral. (In fact, using 3250 rectangles gives a leftsum that
is within 0.00096 of the correct answer.)
8.
Write the equation of the line tangent to the curve
y=3x^{2} + 2x + 1at the point (1,6).
Solution:
The derivative of this function is
y' = 6x + 2, so at the desired point,
the slope is 6+2=8. The tangent line (also known as the first Taylor
polynomial) has the equation
9.
Write the equation of the line tangent to the curve
y^{3}2xy+x^{3}=0at the point (1,1).
Solution:
To find the slope at (1,1), we use implicit differentiation to obtain
3y^{2}y'  2(y + xy') + 3x^{2} = 0. At the point (1,1), this becomes
3y' 2 2y' + 3 = 0, or y'=1. Using this, we get a tangent line
of
10. From physics, we know that the illumination at a point xwhich is provided by a light source at L is proportional to the
intensity of the light at L divided by the square of the distance
between x and L. Suppose that two lights L_{1} and L_{2} are
placed 20 meters apart, and that the intensity of L_{2} is 8 times
the intensity of L_{1}. Where is the point on the line between
L_{1} and L_{2} where the illumination is at a minimum?
Solution:
Let x represent the position of a point on the line from L_{1} to
L_{2}, with L_{1} being at x=0 and L_{2} being at x=20. We need
to write an expression for the illumination at x, and find the value
of x which minimizes it. The illumination provided by L_{1} at xis given by 1/x^{2}, and that provided by L_{2} is
8/(20x)^{2}, so
the total illumination at x is
Taking the derivative, we get
This is only zero when 6x=40, that is, when x=20/3. This is the
only critical point, and it is clearly a minimum because I(0) and
I(20) are both infinite. So the dimmest point is 1/3 of the way
from L_{1} to L_{2}.
11. Find the maximimum and minimum values of the function
f(x)=x^{3} + 3x^{2}  42x  22 on the interval
.
Solution:
First we locate any critical points of the function by solving
f'(x)=0:
Notice that
, which is not in our domain.
So now we should check the values of the function at the relevant
critical point (
) and at the endpoints of
our interval.
and so the maximum of f on [5,2] occurs at
, and
the minimum at x=2.
12. Calculate the area of the region bounded by the graphs of
y=x/2 and x=y^{2}3.
Solution:
The two curves cross at the points (2,1) and (6,3).
This is much easier to do if we integrate with respect to y, because
then top of the rectangles is always given by y=x/2 (that is,
x=2y), and the bottom is the curve x=y^{2}3. So the area is given
by the integral
Notice that if you integrate with respect to x, things are harder
the area is given by
Do you see why?
13.
Write the equation of the parabola which best approximates at
(that is, the second Taylor polynomial). Use your
polynomial to find approximations of the nonzero solutions to
. (Hint: graph the relevant functions for
to make sure your answers make sense. The
fact that
is helpful.)
Solution:
Since
and
, the second Taylor
polynomial at is
Using the quadratic formula to solve where
P_{2}(x)=x/3 gives
, or about 0.694 and
2.114. The solution we want is clearly 2.114, and because of
the symmetry, the other nonzero solution is near 2.114.
14.
Use Newton's method to determine the nonzero solutions of
to within 0.000005. You may either use your answer to the
previous problem as x_{0}, or use x_{0}=2.
Solution:
In this case, the Newton iteration is
Starting with x_{0}=2.114, we get
x_{1} = 2.292061527,
x_{2}=2.278928744,
x_{3}=2.278862662, and
x_{4}=2.278862660. So the
desired result is
.
15. Compute the following limits. Distinguish between ,
, and ``does not exist''.
Solution:
 (a)

does not exist, since
the tangent is periodic with period .
 (b)

 (c)

 (d)

:
As ,
the numerator tends towards 18, and the
denominator tends towards 0. However, since x>3 means the
denominator is positive and x<3 means it is negative, we have
and
.
Hence the
twosided limit does not exist.
 (e)

,
since
e^{x} tends to zero faster than any polynomial grows.
 (f)

(see
part (d)).
16.
The figure below is the graph of a function f(x). Use it to sketch
the graph of f'(x) and the graph of
.
Solution:
17. For each differential equation on the left, indicate which
function on the right is a solution.
(a) y'=5y 

(1) y=x^{2} 
(b) y'=y(y1) 

(2) 
(c)
x^{2} y'' + 2x y' = 1 

(3)
y=(1+e^{x})^{1} 
(d)
yy'' = xy' 

(4) y=e^{5x} 
Solution:
 (a)
 The solution to y'=5y is (4): y=e^{5x}, because for
this function,
y' = 5e^{5x}, which is 5y.
 (b)
 (3) is the solution to
y' = y(y1):
.
On the other side, we have
,
so they are
the equal.
 (c)
 The answer is (2): For ,
we have y'=1/x and
y''=1/x^{2}. Plugging into the expression
x^{2} y'' + 2x y', we get
,
so it satisfies
the equation.
 (d)
 Of course the only choice left is (1), but let's check
anyway. Here we have y=x^{2}, so y'=2x and y''=2. Thus,
yy''=
2x^{2}, and xy'=2x^{2}, so the equation is satisfied.
18.
A coffee filter has the shape of an inverted cone. Water drains from
it at a constant rate of
. When the depth is , the
water level drops at a rate of
. What is the ratio of the
height of the cone to its radius? You may find it useful to recall that
the volume of a cone of radius r and height h is
.
Solution:
Let V(t) denote the volume of water in the cone at time t, and
h(t) be the depth of the water, with r(t) being the radius at
height h. Since the water is in a cone, the ratio r/h is always
constant; let us denote this constant by c, and it is 1/c we need
to determine.
We are told that water drains out at a rate of
 this
says that V'(t)=10. Further, we know that when h(t)=8,
h'(t)=2. Finally, the volume of water at any given time is
.
Note that if we just differentiate the expression for the volume, we
will get something involving both r(t) and r'(t), but we don't
know anything about either of those. However, if we use that
r(t)/h(t)=c, we can rewrite the expression for V(t) not to involve
r(t) at all:
Differentiating and remembering to use the chain rule, we get
, and plugging in gives
. Solving for c yields
19.
Let be the parametric curve given by
What is the slope of at the point (4,2), when t=4?
Solution:
We want to compute
at (4,2) remember that
.
In our case,
or, when t=4, we have
20.
An angle is known to vary periodically with time, in such a
way that its rate of change is proportional to the product of itself
and the cosine of the time t. Write a differential equation which
expresses this relationship. Show that
is a
solution to the differential equation. If you know that
and
, what is the equation for ?
Solution:
This relationship can be written as
The function
is a solution, since
.
If we know
, then we have
2 = ce^{0} = c. Using
, we have
. So
. Taking logarithms, we get
. Thus
Scott Sutherland
19981211