# Math

Topics: Derivative, Trigraph, Differential equation Pages: 29 (3845 words) Published: October 25, 2012
Advanced Mathematics II Homework 1 & 2 (2011)
Homework 1
Sect11-1 Q22 A family paid \$40,000 cash for a house. Fifteen years later, they sold the house for \$100,000. If interest is compounded continuously, what annual nominal rate of interest did the original \$40,000 investment earn?

Solution. Using the continuous compound interest formula we have 100000 = 40000e15r
ln(2.5) = 15r
r = 0.061
where we have put t = 15, P = 40000 and A = 100000. Thus, the annual nominal rate should be 6.1% of interest for the investment.
Sec11-2 Q14 Find the derivative of y = 5e−x − 6ex .
Solution.

dy
= −5e−x − 6ex .
dx

Sec11-2 Q32 Find the derivative of
f (x) =

x+1
ex

and simplify.
Solution.
d
d
dy ex dx (x + 1) − (x + 1) dx ex
f (x) =
=
dx
(ex )2
ex − (x + 1)ex
=
e2x
−x
= −xe .

Sec11-2 Q64
by

The resale value R (in dollars) of a company car after t years is estimated to be given R(t ) = 20, 000e−0.15t .

What is the rate of depreciation (in dollars per year) after 1 year? 2 years? 3 years? Solution. The rate of depreciation after t years is the derivative of R with respect to t : R (t ) =

dR
= −3000e−0.15t .
dt

The rate of depreciation of the car after the ﬁrst, second and third years are: R (1) = −3000e−0.15 = −2582 dollars,
R (2) = −3000e−0.3 = −2222 dollars,
R (3) = −3000e−0.45 = −1913 dollars.
1

Sec11-3 Q38 Find the equation of the line tangent to the graph of y = f (x); f (x) = ln(2 − x) at x = 1.
Solution. By using the chain rule, we have

dy d ln(2 − x
=
dx
dx √

d ln(2 − x) d(2 − x)

=
d(2 − x)
dx
1
11
√ −√
=
2− x
2x
1
√.
=
2(x − 2 x)
Evaluate the slope when x = 1,
1
dy
=
= −0.5
dx x=1 2(1 − 2)
The y coordinate is 0 when x = 1. The point-slope form gives the required equation of the line tangent to f (x):
y = −0.5(x − 1)
at x = 1.
Sec11-3 Q54 Find the absolute maximum value of f (x) = ln(x2 e−x ). Solution. Differentiating f (x) with respect to x gives
f (x) =
=
=
=

d ln(x2 e−x )
dx
d ln(x2 e−x ) d(x2 e−x )
d(x2 e−x )
dx
1
(2xe−x − x2 e−x )
2 − e−x
x
2(1 − x)
.
x

For the local maximum of f (x), putting f (x) = 0 gives x = 1. The second derivative of f (x) with respect to x is
2
f (x) = − 2 .
x
Thus, f (1) = −2 < 0 has veriﬁed the existence of the local maximum, it is f (1) = ln(1e−1 ) = −1.
Since x can be any value except x = 0, i.e. the domain of f (x) is open. Therefore f (1) = −1 at x = 1 is the absolute maximum.
Sec11-3 Q70 On a national tour of a rock band, the demand for T-shirts is given by p = 15 − 4 ln x,

1 ≤ x ≤ 40

where x is the number of T-shirts (in thousands) that can be sold during a single concert at a price of \$ p. If the shirts cost the band \$5 each, how should they be priced in order to maximize the proﬁt per concert?

2

Solution. Let the cost function of the T-shirts be C(x) = ax. Here we ignore the ﬁxed cost term since the T-shirts are supposed not to be manufactured by the band itself. They simply need to order a number of T-shirts that they require. When x = 1, C(1) = a = \$5. The Revenue function R is R(x) = xp = 15x − 4x ln x. The proﬁt function P is P(x) = R(x) − C(x) = 15x − 4x ln x − 5x = 10x − 4x ln x. Differentiating P(x) with respect to x gives

P (x) = 6 − 4 ln x.
Put it equal to zero for the extrema, we get
6 = 4 ln x

x = e1.5 = 4.482 thousands of T-shirts.

The second derivative of P(x) clearly is negative:
P (x) = −4

1
x

< 0.

Thus, 4482 T-shirts have to be sold in order to maximize the proﬁt per concert. Substitute x = 4.482 into the price function, we obtain the price per T-shirt:
15 − 4 ln(4.482) = 9 dollars each.
Sec11-3 Q84 A mathematical model for the average of a group of people learning to type is given by
N (t ) = 10 + 6 ln t , t ≥ 1
where N (t ) is the number of words per minute typed after t hours of instruction and practice (2 hours per day, 5 days per week). What is the rate of learning after 10 hours of instruction and practice?...