Written Homework 8 Solutions
MATH 1310 - CSM
Assignment: pp 164-166; problems: 1 - 4, 8, 11, and 13, pp 172-175; problems: 1 a,c,d,g,h,j,
3, 6, and 9.
PP 164-166
1. Use the chain rule to find dy/dx, when y is given as a function of x in the following way.
Solutions: Note that if y(x) = y(u(x)) then dy/dx = y 0 (u) · u0 (x) by the chain rule.
(a) y 0 = 5 and u0 = −7, so dy/dx = 5 · (−7) = −35
(b) y 0 = cos(u) and u0 = −7, so dy/dx = cos(4 − 7x) · (−7) = −7 cos(4 − 7x)
(c) y 0 = sec2 (u) and u0 = 3x2 , so dy/dx = sec2 (x3 ) · (3x2 ) = 3x2 sec2 (x3 )
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(d) y 0 = ln(10) 10u and u0 = 2x, so dy/dx = ln(10) 10x · (2x)
(e) y 0 = 4u3 and u0 = 3x2 , so dy/dx = 4(x3 + 5)3 · (3x2 ) = 12x2 (x3 + 5)3
2. Find the derivatives of the following functions.
Solutions:
(a) F 0 (x) = 5(9x + 6x3 )4 · (9 + 18x2 )
1
4w
(b) G0 (w) = (4w2 + 1)−1/2 · 8w = √
2
4w2 + 1
1
12w(4w2 + 1)2
(c) S 0 (w) = (4w2 + 1)−1/2 · 3(4w2 + 1)2 · (8w) = √
2
4w2 + 1
1
(d) R0 (x) = (−1)(1 − x)−2 · (−1) =
(1 − x)2
1
−1
(e) D0 (z) = 3 sec2 ( ) · ( 2 )
z
z
(f) dog0 (w) = 2 sin(w3 + 1) · cos(w3 + 1) · (3w2 )
(g) pig0 (t) = − sin(2t ) · ln(2) 2t
(h) wombat0 (x) = ln(5) 51/x · (
−1
)
x2
3. If h(x) = (f (x))6 where f is a function satisfying f (93) = 2, f 0 (93) = −4 what is h0 (93)?
Solution: If h(x) = (f (x))6 then by the chain rule, h0 (x) = 6(f (x))5 · f 0 (x),
Hence by plugging in x = 93 we get h0 (93) = 6(f (93))5 f 0 (93) = (6)(25 )(−4) = −768.
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4. If H(x) = F (x2 − 4x + 2) where F is some function satisfying F 0 (2) = 3, what is H 0 (4)?
Solution: If H(x) = F (x2 − 4x + 2), then by the chain rule we get
H 0 (x) = F 0 (x2 − 4x + 2) · (2x − 4). Then by plugging 4 in for x we get
H 0 (4) = F 0 (42 − (4 · 4) + 2) · ((2 · 4) − 4) = F 0 (2) · 4 = 3 · 4 = 12.
8. Let f (t) = t2 + 2t and g(t) = 5t3 − 3. Determine all of the following:
Solutions:
f 0 (t) = 2t + 2
g 0 (t) = 15t2
g(f (t)) = 5(t2 + 2t)3 − 3
f (g(t)) = (5t3 − 3)2 + 2(5t3 − 3)
g 0 (f (t)) = 15(t2 + 2t)2
f 0 (g(t)) = 2(5t3 − 3) + 2
(f (g(t)))0 = 2(5t3 − 3)(15t2 ) + 2(15t2 )
(g(f (t)))0 = 15(t2 + 2t)2 (2t + 2)
√
11. (a) Write the microscope equation for y = sin x at x = 1.
Solution: Our book says that if y = f (x) is locally linear then at x = a we can write:
√
4y ≈ f 0 (a) · 4x. Since f (x) = y = sin x is locally linear at x = 1, we can write
√
1
4y ≈ f 0 (1) · 4x where f 0 (x) = cos x · √ , so f 0 (1) = 0.27. Hence the microscope
2 x
√
equation for y = sin x at x = 1 is 4y ≈ 0.27 · 4x.
(b) Using the microscope equation, estimate sin
√
√
1.05 and sin 0.9.
√
Solution: In (a) we saw that 4y ≈ 0.27 · 4x, so to estimate sin 1.05 we can use
this microscope equation. We see that if we let 4x = 0.05, then
y(1.05) ≈ y(1) + 4y
√
= sin 1 + (0.27 · 0.05)
= 0.84 + 0.01
= 0.85.
Similarly an estimation of sin
√
0.9 with 4x = −0.1 is y(0.9) ≈ 0.81.
13. This problem was done on the team homework.
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PP 172-175
1. Use differentiation formulas to find the partial derivatives of the following functions.
Solutions:
(a) fx (x, y) = 2xy and fy (x, y) = x2 .
1
1
(c) fx (x, y) = 2xy + 15x2 − √
and fy (x, y) = x2 − √
.
2 x+y
2 x+y
(d) fx (x, y) = ln 10 · 10xy · y and fy (x, y) = ln 10 · 10xy · x.
x2
2x
− 2x · sin y and fy (x, y) = −51 4 − x2 cos y.
3
y
y
v
5
u
5
(h) fu (u, v) = − 2 and fv (u, v) = − 2 .
5 u v
5 uv
(j) fx (x, y) = tan y and fy (x, y) = x sec2 y.
(g) fx (x, y) = 17
3. The volume V of a given quantity of gas is a function of the temperature T (in degrees
Kelvin) and the pressure P . In a so-called ideal gas the functional relationship between
volume and pressure is given by a particularly simple rule called the ideal gas law:
V (T, P ) = R
T
,
P
where R is a constant.
(a) Find formulas for the partial derivatives VT (T, P ) and VP (T, P ).
Solution: VT (T, P ) =
R
T
and VP (T, P ) = −R 2 .
P
P
(b) For a particular quality of an ideal gas called a mole, the value of R can be expressed
as 8.3 x 103 newton-meters per degree Kelvin. (The newton is the unit of force in the
meter-kilogram-second (or MKS) system of units. Check that the units in the ideal
gas law are consistent if V is measured in cubic meters, T in degrees Kelvin, and P
in newtons per square meter.
Solution: This is easy to check.
(c) Suppose a mole of gas at 350 degrees Kelvin is under a pressure of 20 newtons per
square meter. If the temperature of the gas increased by 10 degrees Kelvin and the
volume increased by 1 cubic meter, will the pressure increase or degrease? By about
how much?
Solution: Since we have an equation for V in terms of T and P and we know
R = 8300, we can find the initial volume, V0 using T0 = 350 and P0 = 20, and we
350
get V0 (350, 20) = (8300)
= 145, 250. Next we’re told that the temperature has
20
increased by 10, so T1 = 360 and the volume increased by 1, so V1 (T, P ) = 145, 251.
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Using our same equation for volume, we can solve for P1 :
V1 (360, P1 ) = 145, 251 = (8300)
360
P1
which implies that P1 ≈ 20.57. So we can conclude that the pressure increased by
0.57 newtons per square meter.
6. This problem was not graded.
9. Let P (K, L) represent the monthly profit, in thousands of dollars, of a company that
produces a product using capital whose monthly cost is K thousand dollars and labor whose
monthly cost is L thousand dollars. The current levels of expense for capital and labor
are K = 23.5 and L = 39.0. Suppose now that the company managers have determined
∂P
(23.5, 39.0) = −0.12,
∂K
∂P
(23.5, 39.0) = −0.20.
∂L
(a) Estimate what happens to the monthly profit if monthly capital expenses increase to
$24,000.
Solution: Since we want to estimate what happens to the monthly profit, we should
use the full microscope equation. Let’s set it up: we’re given that Knew = 24, so
4K = Knew − Kold = 24 − 23.5 = 0.5 while 4L = 0 since L is not changing. Thus
our full microscope equation looks like:
4P = PK (K, L) · 4K + PL (K, L) · 4L
Now we plug in:
4P = (−.12)(0.5) + (−.20)(0) = −0.6
Hence the profit decreased by $60 per month.
(b) Each typical person added to the work force increases the monthly labor expense by
$1,500. Estimate what happens to the monthly profit if one more person is added to
the work force. What, therefore, is the rate of change of profit, in thousands of
dollars per person? Is the rate positive or negative?
Solution: Now 4L = 1.5 since each person added to the workforce increases
monthly labor expenses by $1500, while 4K = 0. We use the same microscope
equation as before:
4P = PK (K, L) · 4K + PL (K, L) · 4L
Now we plug in:
4P = (−.12)(0) + (−.20)(1.5) = −0.3
Hence the profit decreased by $300 per month.
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(c) Suppose managers respond to increased demand for the product by adding three
workers to the labor force. What does that do to the monthly profit? If the
managers want to keep the profit level unchanged, they could try to alter capital
expenses. What change in K would leave profit unchanged after the three workers
are added? (This is called a trade-off ).
Solution: If the managers add three workers to the work force, then
Lnew = 39 + 3 · 1.5 = 43.5 and so 4L = 4.5, while 4K = 0. Using the same
microscope equation, we see that 4P = 0 + (−.2)(4.5) = 0.9. So the profit
decreased by $900.
We could also use part (b) since we figured that that the profit decreased
by $300 with the addition of one worker, so if we add three workers, we would
expect the profit to decrease by $300 · 3 = $900.
To determine the trade-off, we use the same microscope equation, set
4P = 0, plug in for the partial derivatives and 4L = 4.5 and then solve for 4K:
4P = PK (K, L) · 4K + PL (K, L) · 4L
0 = (−.12) · 4K + (−.20)(4.5)
0 = (−.12) · 4K − 0.9
0.9 = (−.12) · 4K
4K = −7.5
So we need capital expenses to decrease by $7500 in order to keep the profit level
unchanged.
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