Question 3

 

A Calculator MAY be used to help answer this question.

Be certain to limit yourself to the four permissable calculator functions.

 

Sea grass grows on a lake. The rate of growth of the grass is , where k is a constant.

 

(a) Find an expression for G, the amount of grass in the lake (in tons), in terms of t, the number of years, if the amount of grass is 100 tons initially, and 120 tons after one year.

 

(b) In how many years will the amount of grass available be 300 tons?

 

(c) If fish are now introduced into the lake and consume a consistent 80 tons/year of sea grass, how long will it take for the lake to be completely free of sea grass?

 

Solution

 

a.) By looking at the equation (dG/dt) = kG which represents the rate at which the grass grows on the lake, we can algebraically manipulate it to find the expression that represents G, the amount of grass in the lake with respect to time.

 

(1/G)dG = (k)dt <-- split the operator and move each variable on opposite sides of the equation

S(1/G)dG = S(k)dt <-- antidifferentiate both sides of the equation

ln|G| + c = kt + c

ln|G| = kt + C' <-- combine both constants to form another constant "C' "

G = e^(kt + C') <-- "a natural logarithm is an exponent with base e"

G = (e^kt)(e^C') <-- by the laws of powers

G = Ce^kt <-- e^c is nothing more than another constant, some number, "C"

 

Our work is not complete here, as we are required to find the missing constants "C" and "k". We can do this by inputting the coordinates given in the question above: (0,100), (1,120).

 

100 = Ce^k(0) <-- input the coordinate which corresponds to a 0 "t" value to eliminate the "k" constant and solve "C"

100 = C(1) <-- any value with an exponent equal to 0 evaluates as a value of 1

100 = C <-- the constant "C" is equal to 100

 

120 = 100e^k(1) <-- input the second coordinate pair, as well as the now known value of "C"

1.2 = e^k <-- divide both sides by 100

ln(1.2) = k <-- take the natural logarithm of both sides to find the expression of the exact "k"

 

Now, placing all of the values together, we result with the equation that expresses the amount of grass in the lake, with respect to time.

G = 100e^(ln(1.2))t

 

b.) In this part of the question, we are given a "G" value and are expected to solve for the "t" value: (t,300). Simply here, we input this value, and solve for "t", either algebraically or graphically. Algebraically:

300 = 100e^(ln(1.2))t

3 = e^(ln(1.2))t <-- divide both sides by 100

ln(3) = (ln(1.2))t <-- take the natural logarithm of both sides

t = ln(3) / ln(1.2) <-- divide both sides by ln(1.2) to isolate "t"

t = 6.0257 years <-- solution, with units, approximated to the nearest fourth placed decimal

 

Sentence Answer: In approximately 6.0257 years, the amount of grass will be at 300 tons.

 

c.) By carefully thinking through the question, we find that when the amount of grass consumed by the fish equals to the amount of grass on the lake, then there will be no grass left. Since the fish consume the grass at a constant rate of 80, we can antidifferentiate this to find the expression which represents the amount of grass consumed.

 

f'(t) = 80 <-- rate equation

f(t) = S(80)dt <-- antidifferentiate

f(t) = 80t + c

0 = 80(0) + c <-- when time is 0, we can assume that no grass is consumed (you do not need to assume that c=0, 

0 = c <-- the constant "c" equals 0                                                      the question says that it is a linear

f(t) = 80t                                                                                        function, 80 tons a year is eaten by fish)

 

Now with this new expression, we form an equation to find at what time "t" does the amount of grass consumed by the fish equals to the amount of grass present on the lake:

 

f(t) = G(t)

80t = 100e^(ln(1.2))t

0 = 100e^(ln(1.2))t - 80t <-- form an expression equal to 0 so that the solution can be found graphically

t = 12.7287 years <-- roots of this equation are positioned at that point in time, in years

 

Sentence Answer: In approximately 12.7287 years, the fish will consume all of the grass on the lake.

 

All corrections are in red.