Math
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) =
=
=
=
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) =
=
=
=