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) =

=

=

=