Download HW #3 Solutions

MA 111.1 – SOLUTIONS to Hand-In Homework #3 – 4/13/12
1. A 15-foot ladder is leaning against a wall. The bottom of the ladder is being pushed toward
the wall at a rate of ¼ foot per second (which means the distance from the bottom of the
ladder to the wall is decreasing). If initially the bottom of the ladder is 10 feet away from the
wall, how fast is the top of the ladder moving up the wall after 12 seconds of
pushing?
Solution to #1: Let () = distance from bottom of ladder to wall, and let () =
distance from top of ladder to floor. See diagram at right. We are given that
(0) = 10 and ′() = −1/4 (negative because () is decreasing), and we are
asked to compute ′(12). From the diagram we can see that () + () =
15 = 225. Using implicit differentiation, we find that 2()′() + 2()′() = 0, and thus
′() = −
()′()
()
. Since ′() = −1/4, we know that (12) = 10 −
= 7. By the Pythagorean
Theorem, (12) = √225 − 49 = √176 = 4√11 ≈ 13.2665.Plugging all this into the equation
for ′() and solving gives ′(12) = −
is positive because () is increasing.
()′()
()
/
= − √ = √
≈ 0.1319ft/sec. The answer
2. The population of a frog species on an island has been determined to grow at a rate
proportional to the number of frogs present. If the initial population of frogs is observed to be
500, and after one month the population is 550, find a formula for the number of frogs after t
months. Use this formula to determine how long it will take for the frog population to double.
Give your answers as decimal approximations rounded to the 4th decimal place.
Solution to #2: Because we are told the rate of growth is proportional to the population, we
know that the formula for !(), the number of frogs in the population after t months, is given by
the formula !() = !(0)" # = 500" # . We can determine the value of k by using the given
(()
information that !(1) = 550. Thus, 500" #∙ = 550, and solving for k yields % = ln ()) =
ln 1.1 ≈ 0.0953. Putting this back into the equation for P(t) gives !() = 500" (*+ .) =
500(1.1 ). To determine how long it takes for the population to double, we set P(t) equal to
*+ 1000 and solve for t: 1000 = 500" (*+ .) ⟹ 2 = " (-. .) ⟹ ln 2 =(/0 1.1) ⟹ = *+ . ≈
7.2725 months. So it will take a little more than 7¼ months to double the population.
3. Consider the function, 1(2) = (23 − 4)5 , which is continuous everywhere, on the closed
interval [−5, 8]. Using the methods covered in the slides from 3/29/12, determine the
absolute maximum and absolute minimum of f(x) on this interval, and also indicate all xvalues at which these two extreme values occur.
Solution to #3: We are given a continuous function on a closed, bounded interval, so the
Extreme Value Theorem says that () achieves both an absolute maximum and an absolute
minimum on this interval. We established that the absolute extrema can occur only at a critical
number or an endpoint of the interval. So we determine the critical points, and then examine the
function values at these points and at the endpoints. By the Chain Rule, we see that ′() =
8 ; ( − 1), so the critical numbers of () are 0, 1, and −1. All three of these numbers are in
the interval [−2, 3], so we evaluate () at = 0, 1, −1, −2,and 3: (0) = 1, (1) = (−1) =
0, (−2) = 225, (3) = 6400. Thus the absolute minimum of () on[−2, 3] is 0, which occurs
twice, at = 1and = −1, and the absolute maximum is 6400, which occurs at = 3.
4. Consider the function, 1(2) = 23/8 (25 − <), which is continuous on the whole real line.
a. Determine the exact values of all critical numbers of ().
b. For each critical number = found in part (a), apply the Second Derivative Test, if
possible, to determine if 1(=) is a local max, local min, or neither.
c. For each critical number = for which the Second Derivative Test does not give an
answer to the question in part (b), explain why that is the case, and then use either the
First Derivative Test or the Function Value Test to determine if f(c) is a local max,
local min, or neither.
Solution to #4:
a. We can rewrite () = /; ( − 5) = )⁄; − 5 ⁄; , and so ′() =
factor to get ′() =
)
;
⁄; ( − 2).
)
;
⁄; −
)
Thus the critical numbers of () are = 0, √2 ≈ 1.414,and = −√2 ≈ −1.414.
b. From
)
@ A⁄B
′() =
)
;
⁄; −
)
;
⁄; ,
we
get
′′() =
)
@
⁄; −
)
@
⁄; = )
@
;
⁄; . We
⁄; −
)
@ A⁄B
=
(7 − 2). The Second Derivative Test says that if C is a critical number, and ′′(C) < 0,
then (C) is a local maximum, and if ′′(C) > 0, then (C) is a local minimum. If either ′′(C)
is undefined or ′′(C) = 0, the Second Derivative Test fails to give any information. In this
problem ′′(0) is undefined, but, using the formula ′′() =
)
@ A⁄B
(7 − 2), it’s easy to see
that both ′′F√2G > 0 and ′′F−√2G > 0. Thus F√2G and F−√2G are local minimums and
B
they have the same value: F√2G = F−√2G = −3√4 ≈ −4.7622.
c. We can apply either the Function Value Test or the First Derivative Test at the critical
number C = 0. I’ll show both answers (you only needed to do one or the other).
For the Function Value Test we compare the value of (0) with values of the function a
little to the left and a little to the right of C = 0, say at = −1 and = 1 (which are close
enough to 0 to exclude the other critical numbers). From the formula, () = /; ( − 5),
we see that (0) = 0, (1) = −4, and (−1) = −4, so we conclude by the Function Value
Test that (0) = 0 is a local maximum.
For the First Derivative Test, we look at the sign of ′() a little to the left of 0 and a little
)
to the right of 0. Again, I’ll use = 1 and = −1. From the formula, ′ () = ; ⁄; ( − 2),
we see that ′(−1) > 0 and ′(1) < 0. So () is increasing to the left of = 0 and
decreasing to the right of = 0, and hence () has a local maximum at = 0.
5. Using the “number lines” method discussed today in class, draw a good sketch of the graph of
the function, 1(2) = 4H28 − I2<. Be sure to show all the work outlined in the steps given on
Slide #3 from 4/5/12 (Thursday’s slides). I have updated that slide to reflect the steps we did
in class today.
Solution to #5: First we do a number line for ′(). Facto to get ′() = 30 − 30 =
30 (1 − ), so ′() = 0 at = 0, 1, and −1. We put these values on a number line and test
the sign of ′() in between. The number line for ′() is shown below. We also must do a
number line for ′′() = 60 − 120 ; = 60(1 − 2 ). Thus ′′() = 0 at = 0,
and
√
√
≈ 0.7071,
≈ −0.7071. The number lines for ′() and ′′() appear below, and below them is the
corresponding information about where () is increasing/decreasing and concave up/concave
down. Next, we find the - and -intercepts of () = 10 ; − 6 ( = 2 ; (5 − 3 ). The intercept (also an -intercept) is (0, 0). In addition JK5⁄3 , 0L and J−K5⁄3 , 0L are the other x-
intercepts (note: K5⁄3 ≈ 1.2910). Finally, we should compute the function values at the critical
numbers:
(−1) = −4, (1) = 4, (1M√2) =
√
≈ 2.4749,and (−1M√2) =
√
≈ −2.4749.
The graph (done in Mathematica) appears below.
10
5
-1.5
-1.0
0.5
-0.5
-5
-10
1.0
1.5