Solutions to HW Set 5
1.7.16) Find the first and second derivatives of v = 2t2 + 3t + 11.
dv
= 4t
dt
d2 v
=4
dt2
+3
1.7.26.) Compute
d dy
( ) |x=1 ,
dx dx
where y = x3 + 2x − 11
dy
dx
= 3x2 + 2
d dy
d dy
( dx ) = 6x, so, plugging in 1 for x, we get dx
( dx ) |x=1 = 6
so dx
√
1.7.36.) Suppose that s = 7x2 y z. Find
a.)
ds
dx
b.)
ds
dy
d2 s
dx2
√
= 14xy z, so taking the derivative with respect to x again, we get
d2 s
dx2
√
= 14y z.
d2 s
dy 2
√
= 7x2 z, which is constant with respect to y, so
ds
c.) dz
ds
= 7x2 y ∗
dz
d2 s
dy 2
=0
1
√
2 z
1.7.38) Estimate the cost of manufacturing 51 bicycles per day if C(50) = 5000 and C 0 (50) =
45, where C(x) is the cost of manufacturing x bicycles per day.
We can approximate the function C(x) for values near x = 50 by using the line tangent
to the graph at x = 50. We know (50, 5000) is on the tangent line, and we know the slope of
the tangent line is C 0 (50) = 45. So the equation of the tangent line is y = 45x + b. Pluggin
in the point (50, 5000), we get 5000 = 45 ∗ 50 + b. Solving for b, we get b = 2750, so the tangent line is y = 45x + 22750. This gives us the approximation C(51) ≈ 45 ∗ 51 + 2750 = 5045.
1.7.39)Suppose the revenue from producing x units of a product is given by R(x) = 3x−.01x2
dollars.
a.) Find the marginal revenue at a production level of 20.
Marginal revenue of producing a units is the quantity R0 (a). For us, R0 (x) = 3 − .02x,
so at a production level of 20, we have R0 (20) = 3 − .02(20) = 2.6dollarsperunit
b.)Find the production levels where the revenue is $200.
We want to find the x such that R(x) = 3x − .01x2 = 200. This gives us the quadratic
equation .01x2 − 3x + 200 = 0. We multiply through by 100 to get x2 − 300x + 20000 = 0.
The quadratic formula gives us as solutions x =
200 units.
300±100
,
2
so the production levels are 100 or
1.7.40) Let P (x) be the profit from producing and selling x units of goods. Match each
question with the proper solution.
Question A: What is the profit from producing 1000 units of goods?
Solution d.) Compute P (1000)
Question B: At what level of production will the marginal profit be 1000 dollars?
Solution b.) Find a value of a for which P 0 (a) = 1000.
Question C: What is the marginal profit from producing 1000 units of goods?
Solution a.) Compute P 0 (1000)
Question D: For what level of production will the profit be 1000 dollars?
Solution c.) Set P (x) = 1000 and solve for x.
1.8.8)Suppose that t days after the end of the advertising, the daily sales are f (t) =
−3t2 +32t+100 units. What is the average rate of growth in sales during the fourth day, that
is, from time t = 3 to t = 4? At what (instantaneous) rate are the sales changing when t = 2?
2
2
(3)
+32(3)+100)
The average rate of change between t = 3 and t = 4 is f (4)−f
= −3(4) +32(4)+100−(−3(3)
=
4−3
1
−48 + 128 + 100 + 27 − 96 − 100 = 11
The instantaneous rate of change at t = 2 is just f 0 (2). f 0 (t) = −6t + 32, so f 0 (2) =
−12 + 32 = 20.
7
1.8.12)Suppose the position of a car at time t (in hours)is given by s(t) = 50t − t+1
, where
the position is measured in kilometers. Find the velocity and acceleration of the car at t = 0.
7
The velocity is the derivative of position, so v(t) = s0 (t) = 50 + (t+1)
Thus, at t = 0
2.
7
the velocity is v(0) = 50 + (0+1)2 = 57.
−14
−14
Acceleration is the derivative of velocity, so a(t) = v 0 (t) = (t+1)
3 . Thus, a(0) = (0+1)3 = −14
1.8.18)A car is traveling from New York to Boston and is partway between the two cities.
Let s(t) be the distance from New York during the next minute. Match each behavior with
the corresponding graph of s(t) in Fig. 8.
a.)The car travels at steady speed: graph b
b.)The car is stopped: graph c
c.)The car is backing up: graph d
d.)The car is accelerating: graph a
e.)The car is decelerating: graph e
1.8.22)Suppose 5mg of a drug is injected into the bloodstream. Let f (t) be the amount
present in the bloodstream after t hours. Interpret f (3) = 2 and f 0 (3) = −.5. Estimate the
number of milligrams of the drug in the bloodstream after 3.5 hours.
f (3) = 2 means there are 2 mg of the drug present after 3 hours. f 0 (3) = −.5 means
the amount of drug present is changing at a rate of -.5 mg per hour after 3 hours.
We estimate f (3.5) by taking a linear approximation to f at 3. We know the slope of f
at 3 is m = −.5 so the approximation is y = −.5t + b. We further know that (3, 2) is
on the graph of f and on our linear approximation, so we know 2 = −.5(3) + b. Solving
for b, we get b = 7/2 = 3.5. So our approximation at t = 3.5 is f (3.5) ≈ −.5(3.5)+3.5 = 1.75.
Problem 1) A toy rocket is launched from the ground. It accelerates straight upwards at 100 feet/sec/sec for
10 seconds, and then runs out of fuel. What is the maximum height it eventually reaches? How long after
launch does it hit the ground? How fast is it going when it hits the ground?
The rocket’s motion has 2 distinct parts: 1.)The rocket accelerates upward using its engines, 2.)The rocket
is out of fuel, but continues going upward until gravity forces it to stop, and then falls down. We will want
a set of equations to describe its motion in each of these parts.
For the first part, we know that the acceleration a1 (t1 ) is a constant 100 ft/s/s, so we wrtie a1 (t1 ) = 100.
Acceleration is the derivative of velocity, so we know there is a function for the rocket’s velocity during the
first part of its journey such that v10 (t1 ) = a1 (t1 ) = 100. Thinking about this, we see that v1 (t1 ) must be a
function of the form v1 (t1 ) = 100x + v0 for some constant v0 . Since v1 (0) = 100(0) + v0 = v0 and the rocket
starts at rest (i.e. v1 (0) = 0), we get that v0 must be 0, so v1 (t1 ) = 100t1 .
Now, the derivative of position is velocity, so we must have a function d1 (x) that gives the rocket’s distance
above the ground such that d01 (t1 ) = 100t1 . Clearly, this must be a function of the form d1 (t1 ) = 50t21 + d0 ,
for some constant d0 . As above, d1 (0) = 0 = 50(0) + d0 = d0 , so d1 (t1 ) = 50t21 .
These equations allow us to calculate all the quantities for the first part of the trip. However, the max
height will be achieved while we are in the second part of the trip, so we have more work to do.
As before, we start out with an equation for acceleration, a2 (t2 ) = −32, since -32 ft/s/s is the acceleration due to gravity, and t2 is taken to be the time, in seconds, since the rocket has run out of fuel. Using
similar methods to part 1, we get a function for velocity, v2 (t2 ) = −32t2 + v00 , for some other constant v00 .
Now, v00 is the velocity at time t2 = 0, which is the time when the rocket runs out of fuel, i.e. time t1 = 10.
So v00 = v1 (10) = 100 ∗ 10 = 1000, and v2 (t2 ) = −32t2 + 1000.
Then we get an equation for distance above the ground for part 2, d2 (t2 ) = −16t22 + 1000t2 + d00 using the same trick of thinking of the function whose derivative is the formula we have for velocity. Here
d00 is the distance at time t2 = 0 which is the time t1 = 10, so d00 = d1 (10) = 50 ∗ 100 = 5000 and
d2 (t2 ) = −16t22 + 1000t2 + 5000.
Now we can solve for the maximum height, since this will be achieved when the velocity of the rocket
in part 2 becomes 0. The maximum height will be the distance given by d2 (t2 ) at the time t2 when we reach
the top of the arc, stop, and begin falling, so we need to find how long in seconds it takes to reach this
stage. This we can do with our equation for velocity. We know that when we stop moving upwards and are
about to begin falling, the rocket’s velocity must be zero. So we simply solve v2 (t2 ) = 0 for t2 . That is,
−32t2 + 1000 = 0. This gives us t2 = 31.25 as the time when the rocket stops going up, so this is the value
we plug in to d2 (t2 ) to get our maximum height(the rocket can’t go higher than it was when it stopped going
up). Doing the calculations, we get d2 (31.25) = 20625 ft as our maximum height.
Now we want to know with what velocity the rocket hits the ground. We use the equations we have already set up, a2 (t2 ) = −32, v2 (t2 ) = −32t2 + 1000, and d2 (t2 ) = −16t22 + 1000t2 + 5000
We can use these to find how long it takes for the rocket to hit the ground, and use that time to find
the velocity it has when it does. When the rocket hits the ground,
√ it will have d2 (t2 ) = 0 so we solve this
equation for t2 . 0 = −16t22 + 1000t2 + 5000 gives us t2 =
√
1000∓200 33
32
=
√
125∓25 33
.
4
−1000±
1000000−4(5000)(−16)
−32
=
√
−1000± 1320000
−32
=
We take the positive solution since we are dealing with time and the negative
√
33
solution does not make sense, so t2 = 125+25
. This is the number of seconds after we run out of fuel that
4
√
√
33
125+25 33
the rocket hits the ground, so the velocity it has when it hits is v2 ( 125+25
)
=
−32
+ 1000 =
4
4
√
√
−200( 33 + 5) + 1000 = −200 33 − 1000 + 1000 ≈ −1148.91 ft/s.
The total time of the rockets flight is just
the sum of the time spent in each of the 2 parts. Part 1 lasts 10
√
33
seconds by definition, Part 2 lasts 125+25
seconds before the rocket hits the ground. Thus, the total time
4
√
125+25 33
≈ 10 + 67.154 = 77.154 seconds.
is 10 +
4
Problem 2.)Find the equation of the tangent line from (0, −16) to y = x3
We are looking for the line through (0, −16) tangent to y = x3 . This line must be tangent to the curve at a
point, so call this point (x1 , y1 ). Then the slope of the tangent line must be given by the derivative at x1 ,
x3 +16
y1 +16
2
i.e. m = 3x21 . Now, the slope is also given by the slope formula as m = y1 −(−16)
= 1x1 .
x1 −0 . So 3x1 =
x1
Multiplying both sides by x1 yields 3x31 = x31 + 16, so x31 = 8, which means x1 = 2. Thus, (x1 , y1 ) must be
the point (2, 8), so the slope of the tangent line is m = 3(2)2 = 12 and the y-intercept is given as −16, so
the tangent line is given by the equation y = 12x − 16.
Problem 3.)Let C1 be the graph of the curve y = x2 , and let C2 be the graph of the curve y = −2x2 + 4x − 3.
Find the equation of a straight line y = mx + b that is tangent to both C1 and C2. There are two such lines;
find both of them. Hint: each line will intersect C1 and C2 at two different points (x1 , y1 ) and (x2 , y2 ). Try
to find these points.
Say (x1 , y1 ) is on the curve C1, so it satisfies y1 = x21 and (x2 , y2 ) is on the curve C2,
so it satisfies y2 = −2x22 + 4x2 − 3. Since the tangents at these points give the same line, the
derivatives must be equal, i.e. 2x1 = −4x2 +4, so x1 = −2x2 +2. The slope of this line is also
−2x22 +4x2 −3−x21
−y1
given by m = xy22 −x
=
. We convert this to a formula of 1 variable by plugging
x2 −x1
1
−2x22 +4x2 −3−(4x22 −8x2 +4)
−2x22 +4x2 −3−(−2x2 +2)2
=
.
x2 −(−2x2 +2)
3x2 −2
2
2
2
−2x2 +4x2 −3−4x2 +8x2 −4)
−6x2 +12x2 −7
Now, m = −4x2 + 4, so we get an equation −4x2 + 4 =
=
.
3x2 −2
3x2 −2
2
2
We can multiply both sides by 3x2 − 2 to get −12x2 + 20x2 − 8 = −6x2 + 12x2 − 7. This
√ gives
2
us the quadratic equation 6x2 − 8x2 + 1 = 0. We solve this to get x2 = (2/3) ± ( 10/6).
expressing x1 as a function of x2 , i.e. m =
Each solution gives a tangent line.
√
√
+
(2
10/9).
Starting with the solution x2 = (2/3) + √
( 10/6), we get that y2 = (−16/9)
√
Now
√ we know m = −4x2 + 4 = (4/3) − (2 10/3) and we know ((2/3) + ( 10/6), (−16/9) +
line, so we can solve for the equation of the tangent line
(2 10/9))√is on the tangent
√
2 10
4 10
4
−14
y = ( 3 − 3 )x + 9 + ( 9 ).
√
√
Starting with the solution x2 = √
(2/3) − ( 10/6), we get that y2 = (−16/9) − (2 10/9)
and m = −4x2 + 4 = (4/3) + (2√ 10/3). So we √
can do a lot of arithmetic to get that our
2 10
4 10
4
−14
second tangent line is y = ( 3 + 3 )x + 9 − ( 9 ).
Problem 4.) Let C1 be the graph of the curve y = x2 , and let C2 be the graph of the curve y = −2x2 +4x−c,
where c is a constant. For which value(s) of the constant c are there are exactly two different lines tangent to
C1 and C2? For which value(s) of the constant c is there exactly one line tangent to C1 and C2? For which
value(s) of the constant c are there no lines tangent to C1 and C2? Describe these three cases in terms of a
geometric property of C1 and C2 (it may help to graph C1 and C2 in these 3 cases).
We can proceed as we did in Problem 3.) (since the curves are almost the same) until we get to the
−2x22 +4x2 −c−(−2x2 +2)2
−2x22 +4x2 −c−(4x22 −8x2 +4)
m =
=
, where we have the constant c in our expression.
x2 −(−2x2 +2)
3x2 −2
−2x2 +4x −c−4x2 +8x −4)
−6x2 +12x −c−4
2
2
2
2
2
2
We get the equation −4x2 + 4 =
=
and multiply both sides by
3x2 −2
3x2 −2
3x2 − 2. So we get −12x22 + 20x2 − 8 = −6x22 + 12x2 − c − 4, which again will give us a quadratic equation: 6x22√− 8x2 + (4 − c) = 0. We set this up to solve, but the solution will depend on c. We get
8±
64−4(6)(4−c)
. Now, in problem 3.) we got 2 solutions to our quadratic, each of which gave a
x2 =
12
distinct line tangent to both curves. This was because any x2 we get will give a unique pair y2 and m, which
determine a unique line. Thus, the number of lines tangent to both curves is the same as the number of solutions to our quadratic. We therefore need to find the ranges of c that give us different numbers of solutions.
We have exactly 1 solution when c is such that the discriminant (64 − 24(4 − c)) equals 0. Then the ±
will be irrelevant. So we solve 64 − 24(4 − c) = 0 for c. This gives us c = 4/3, so for this value we will
only get 1 line that is tangent to both curves. For c’s greater than this, the discriminant
is greater (the
√
discriminant d = 64 − 96 + c, so it increases when c does). So for c > 4/3, d > 0, so x2 = 8±12 d gives 2 unique
√
solutions, each of which gives a line tangent to both curves. For c < 4/3, d < 0, so x2 = 8±12 d involves
taking the root of a negative number, which means it has no solutions in the real numbers, and so we have
no lines tangent to both curves.
Geometrically, the cases c > 4/3, c = 4/3, c < 4/3 correspond to the curves C1 and C2 not touching,
intersecting at 1 point, or intersecting at 2 points. This is because the vertical distance between the parabolas is smallest at x = 2/3, and is in fact −4/3 + c. So when c = 4/3 this distance becomes 0, and when
c < 4/3 this distance is negative, which means at x = 2/3, C2 is above C1, so the curves will intersect once
on either side of x = 2/3. See the separate pictures for examples of graphs of each of these three cases.