Practice problems
Physics
Exercise 1. A ball is thrown into the air from the top of a hill. The height of the ball is given by the position function
s(t) = −16t2+ 48t+ 160,
wheres is measured in feet andtis measured in seconds. Find the maximum height of the ball and calculate its velocity and acceleration at that point.
Solution. The extrema can be found by second derivative test. The first derivative of f is s(t) =−32t+ 48
and the critical point ist = 48 32 = 3
2. Since
s(t) =−32
is always negative, the critical point is a maximum point, which is a global, since s(t) is a quadratic function. The maximum height of the ball is
s(3/2) =−16·9
4 + 48· 3
2 + 160 =−36 + 72 + 160 = 196.
It is evident that the velocity, which is s(t) must be zero at this time, the ball is stopping for a moment at its maximum height. But we can validate it by substituting t = 3/2 into s(t) which measures the velocity:
s(3/2) =−32· 3
2 + 48 = 0.
The second derivative measures the acceleration, that is constant (due to the gravity of Earth).
The negative value indicates that the ball is being pulled down – toward the earth.
93
Exercise 2. A hot air balloon has a velocity of 50 meter/min and is flying at a constant height of 600 meters. An observer on the ground is watching the balloon’s approach. How fast is the changing of distance between the balloon and the observer when the balloon is 1000 meters from the observer?
Solution. If we denote the distance between the balloon and the observer with r and the horizontal distance from the balloon to the point directly above the observer with x, then by Pythagorean theorem,
r2(t) = 6002+x2(t) where t is measured in minutes.
ground
600
x
r
observer
balloon
We need to find r(t) when r(t) = 1000. Taking the derivative with respect to t of both sides, using the fact the r(t) and x(t) are functions depending ont, we obtain
2r(t)·r(t) = 0 + 2x(t)·x(t).
Substituting r(t) = 1000, we get
2·1000·r(t) = 2x(t)·x(t).
We know, that x(t) = 50 all the time, hence
2000r(t) = 2x(t)·50
From the equation r2(t) = 6002+x2(t),r(t) = 1000 givesx(t) = 800, thus 2000r(t) = 2·800·50 = 80000
r(t) = 40.
The changing of the distance between the balloon and the observer at the specified position is 40 meter/min.
Exercise 3. Research has shown that the number of miles per gallon for a Nissan 200SX automobile is given by the function
M(x) = 2000x
1000 +x2, 10≤x≤100,
wherex is the speed of the car in miles per hour. Determine the speed in the specified interval that yields the greatest fuel economy.
Solution. We have to find the absolute maximum of the function M on the interval [10,100].
1. Evaluating M(x) at the endpoints 10 and 100, we have M(10) = 2000·10
1000 + 100 = 200
11 ≈18.18, M(100) = 2000·100
1000 + 1002 = 2000
10 + 102 = 200
11 ≈18.18.
2. The derivative of M is
M(x) = 2000(1000 +x2)−2000x·2x
(1000 +x2)2 = 2000(1000−x2) (1000 +x2)2 .
Since M is differentiable at all points, we find the critical points by solving the equation M(x) = 0, that is equivalent with
1000−x2 = 0.
This gives two critical points, x=−√
1000 and x=√
1000, but only the last one lies in the interval (10,100). The evaluation ofM(x) at this point gives
M√ 1000
= 2000·√ 1000 1000 + 1000 =√
1000 = 10√
10≈31.62.
3. Hence the absolute maximum value of M(x) is 10√
10, which occurs at the point 10√ 10.
Biology
Exercise 4. The concentration of a certain drug in the bloodthours after an injection is given by the function
C = 3t
54 +t3, 0≤t.
When will the concentration be the highest?
Solution. The function C is differentiable for 0≤t. The derivative is C = 3·(54 +t3)−3t·3t2
(54 +t3)2 = 162−6t3 (54 +t3)2,
which is zero when 162−6t3 = 0, that ist3 = 27. Hence, there is only one critical point,t = 3.
Since
C(2) = 162−6·8
(54 + 8)2 >0 and C(4) = 162−6·64 (54 + 64)2 <0,
theCis increasing before and decreasing aftert= 3. Therefore, the concentration is the highest 3 hours after the injection.
Exercise 5. The height of a sunflower plant follows the logistic growth law, so it can be measured by the function
H(t) = 2et−4
1 +et−4, 0≤t,
where the height of the sunflower is in meters and t is the time, measured in weeks. At what height will the rate of growth be maximal?
Solution. We have to find the maximum place of the derivative function H, which is an inflection point of H, and to evaluate the function H at this point.
H = 2et−4·(1 +et−4)−2et−4·et−4
(1 +et−4)2 = 2et−4 (1 +et−4)2
H= 2et−4·(1 +et−4)2−2et−4·2(1 +et−4)·et−4 (1 +et−4)4
= 2et−4·(1 +et−4)−2et−4·2et−4 (1 +et−4)3
= 2et−4·(1−et−4) (1 +et−4)3
Sinceet−4is a strictly increasing positive function, the only critical point arise from the following equation.
1−et−4 = 0 et−4 = 1 =e0 t−4 = 0
t= 4 The function 1−et−4 is positive before and negative after the point t = 4, hence this is a global maximum place forH, and of course an inflection point for H. Thus, the height of the plant is
H(4) = 2e4−4
1 +e4−4 = 2e0
1 +e0 = 1 meter
when its growth rate is maximal. 4
1
t (week) H (meter)
In economics this point is called the “point of diminishing returns”.
Economics
Exercise 6. A deposit of $10,000 is made in a savings account that pays an annual interest rate of 6%. Find the balance in the account at the end of the 8th year if the interest is compounded continuously. After how many years will the balance be more than double of the original deposit?
Solution. IfP is the amount of deposit,t is the number of years,Ais the balance aftert years and r is the annual interest rate (in decimal form).
• Compounded n times per year: A=P 1 + r
n nt
.
• Compounded continuously: A=P ert.
Now, P = $10,000, t= 8 andr = 0.06. Hence, the balance after continuous compounding is A=P ert= 10,000e0.06·8 = 10,000e0.48 ≈16,160.
To double the deposit, t must satisfy the inequality 10,000e0.06t≥20,000
e0.06t≥2
0.06t≥ln 2 = 0.68314718 t≥11.55.
The balance will be more than double of the original deposit at the end of the 11th year.
Exercise 7. The number of daily sales of a product is given by the function S(x) = 100xe−x2/18+ 50.
The variable x measures the number of days after the start of an advertising campaign. Find the maximum daily sales during the first 18 days of the campaign.
Solution. We have to find the maximum of the function S on the closed interval [0,18].
S(0) = 0
S(18) = 1800e−18+ 50≈50 The derivative of S is
S(x) = 100e−x2/18+ 100xe−x2/18· −x
9 = 100e−x2/18·
1− x2 9
.
Therefore, the critical points are x = −3 and x = 3, but only the last one lies in the interval (0,18). EvaluatingS at this point gives
S(3) = 300e−1/2+ 50 ≈232.
So, 3 days after the start of the campaign we have the maximum daily sales with 232 products.
Supplementary exercises
Physics
Exercise 1. A ball is dropped from a tall building. The function s(t) = 16t2 represents the distance between the original position of the ball and its position after t seconds. Assume no air friction to find the indicated quantities.
(a) Find the distance fallen during the first 3 seconds.
(b) Find the distance fallen during the first 10 seconds.
(c) Find the velocity of the ball at time t= 3 seconds.
(d) Find the velocity of the ball at time t= 10 seconds.
(e) At what time will the velocity be the same as the average velocity over the first 5 seconds?
Solution.
(a) 144 feet (b) 1600 feet
(c) t= 96 feet per seconds (d) t= 320 feet per seconds
(e) 2.5 seconds
Exercise 2. The position s(t) of an object moving on the x-axis is given by the function s(t) = 10− 20
t2+ 1,
where t is measured in seconds. What is the velocity of the object at any time t?
Solution. v(t) = 40t (t2+ 1)2
Exercise 3. Two towns lie on the south side of a river. A pumping station is to be located to serve the two towns. A pipeline will be con-structed from the pumping station to each of the towns along the line connecting the town and the pumping station. Locate the pump-ing station to minimize the amount of pipeline that must be constructed.
5mi
2mi
A P
B 10 mi
Solution.
A P
B 20
7
50 7
Biology
Exercise 4. Medical data predict that the response y in millivolts of a muscle to the amount x of adrenalin injected in cubic centimetres is given by the function
y(x) = x ax+b. What is the rate of change of y with respect to x?
Solution. b (ax+b)2
Exercise 5. A bacterial infection is being treated with a new form of hydrocortisone. Suppose that every week the bacteria population falls to 80% of the population of the previous week. If the initial population is 500, what is the exponential function that describes the population at any time t >0?
Solution. 500·0.8t
Exercise 6. W. Estes was able to estimate the “learning curve” of rats, that is the number of correct selections per minute over a given time period, by the function
r(t) = 13
1 + 25e−0.24t, t≥0,
where r is the predicted number of correct selections per minute and t is the time (min) that the experiment has elapsed.
(a) Show that the learning curve r(t) is always increasing.
(b) What is the limit value ofr(t) as t approaches infinity?
Solution.
(a) Its derivative is
r(t) = 78e−0.24t (1 + 25e−0.24t)2 and this is always positive.
(b) lim
t→∞r(t) = 13
Exercise 7. The size of a population of a bacteria growing in a liquid medium is given by the function
p(t) = 100e0.04t,
wheretis the time measured in months. Draw the graph of the relative rate of increasep(t)/p(t) in the population over the next ten months.
Solution.
10 0.04
t p
Economics
Exercise 8. The daily demand for automobile gasoline as a function of price has been deter-mined to be
D(p) = (p−5)2
2 0.5≤p≤5,
where D is measured in million of litres and pis measured in dollars.
(a) Find the total decrease in demand when the price of gasoline is raised from $1 to $5.
(b) Find the average decrease in demand when the price is raised from $1 to $5.
(c) Find the total decrease in demand when the price is raised from $2 to $5.
(d) Find the average decrease in demand when the price is increased from $2 to $5.
Solution.
(a) 8 litres (b) 2 litres (c) 4.5 litres (d) 1.5 litres Exercise 9. Find the price elasticity E = (p·S)/S for the supply function
S(p) =p2−5 when the price is $4.
Solution. E = 32/11
Exercise 10. The daily revenue R received from selling q units of a certain product is given by the function
R(q) =
400q−q2, 0≤q ≤400.
What is the marginal revenue when sales are 50 units a day?
Solution. R(50) = 3√ 17500 350
Exercise 11. Supertankers off-load oil at a docking facility 4 miles offshore. The nearest refinery is 9 miles east of the shore point nearest the docking facility. A pipeline must be constructed connecting the docking facility with the refinery. The pipeline costs $300,000 per mile if constructed underwater and $200,000 per mile if overland.
A B Refinery
Shore
9 mi Docking facility
4mi
(a) Locate pointB to minimize the cost of the construction.
(b) The cost of underwater construction is expected to increase, whereas the cost of over-land construction is expected to stay constant. At what cost does it become optimal to construct the pipeline directly to pointA?
Solution.
(a) The point B is 8/√
5≈3.58 mi far away from A.
(b) The cost underwater pipeline should be infinite per mile.
Exercise 12. A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
Solution.
200 m 400 m
Area: 80000 m2
Quizzes
Version A
Exercise. A painter is painting a house using a 15 ft long ladder. A dog runs by the ladder dragging a leash that catches the bottom of the ladder and drags it directly away from the house at a rate of 22 ft/sec. Assuming the ladder continues to be pulled away from the wall at this speed, how fast is the top of the ladder moving down the wall when the top is 5 ft from the ground?
Version B
Exercise. A motion of a body on a line is given by the function s(t) = (2−t)√
5−t, 0≤t≤4, where
(a) Find the body’s speed and acceleration at the endpoints of the interval.
(b) When, if ever, during the interval does the body change direction?
Version C
Exercise. A firm can produce between 100 and 1000 pairs of gloves per day. The cost of producing a pair of gloves in dollars is given by
C(x) = 297 + 3.28x+ 0.003x2. The demand function for the gloves in dollars per pair is
D(x) = 7.47 + 321 x .
(a) Find the marginal cost, marginal average cost and marginal profit functions.
(b) Determine the number of pairs that should be produced to minimize the average cost.
(c) Determine the number of pairs that should be produced to maximize the profit.
Solution of Quizzes
Version A
Solution of Exercise.
15 ft x
y
22secft
We can state that x2+y2 = 152. But x and y are the functions of the timet as well. So x(t) = 22.
The question is y(t), ify(t) = 5. Ify= 5, then x=
152−y2 =√
225−25 = √ 200.
Furthermore,
y(t) =
225−x(t)2 y(t) = −x
√225−x2 ·x(t) y(t) = −√
√ 200
225−200 ·22 = −44√
2≈ −62.23.
The speed of the top of the ladder at the observed position is approximately 62.23 ft/sec.
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2 pt
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5 pt
Version B
Solution of Exercise.
(a) Speed:
v(t) = s(t) = −√
5−t− 2−t 2√
5−t
• v(0) =−√ 5− 1
√5
• v(4) = 0 Acceleration:
a(t) =v(t) = 1 2√
5− 1 2 ·
−√
5−t+ 2√2−5−tt 5−t
.
• a(0) =
√5 5 − 1
1√ 5
• a(4) = 3 2
(b) The body changes its direction, where s(t) has extremal point, i.e. v(t) = 0.
−√
5−t− 2−t 2√
5−t = 0
−2(5−t)−(2−t) 2√
5−t = 0
−2(5−t)−(2−t) = 0 3t−12 = 0 t = 4
But this critical point is an endpoint of the interval, so the body does not change its direction within the observed interval.
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2 pt
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5 pt
Version C
Solution of Exercise.
(a) Marginal cost:
M C =C(x) = 3.28 + 0.006x Average cost:
AC = C(x)
x = 297
x + 3.28 + 0.003x Marginal average cost:
M AC =AC(x) = −297
x2 + 0.003 Profit:
Π(x) = xD(x) = 7.47x+ 321 Marginal profit:
MΠ = Π(x) = 7.47
(b) The average cost can be minimal where the marginal average cost is zero.
−297
x2 + 0.003 = 0 297
x2 = 0.003 x2
297 = 1000 3 x2 = 297000
3
x=± 297000
3 =±99000
None of the critical points lies in the interval [100,1000], so the minimal average cost can be obtained in the endpoints. Since
AC(10) = 29.7 + 3.28 + 0.03 > AC(100) = 2.97 + 3.28 + 0.3, the minimal average cost can be reached by producing 100 pairs of gloves.
(c) The profit can be maximal where the marginal profit is zero, but it never happens. So the profit can be maximal producing 10 or 100 pairs of gloves. Since Π is monotone increasing, the maximal profit is
Π(100) = 747 + 321 = 1068.
1 pt
2 pt
3 pt
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5 pt