Question

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

a. Find an expression for ????, the amount of grass in the lake (in tons), in terms of ????, 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?

Answer 1

`(a)\ G(t) = 100 *e^(0.1823t)`

`(b)\ t = 6`

`(c)\ t = 1.7`

Step-by-step explanation:

Given

`G_0 = 100` --- initial

`G(1) = 120` --- after 1 year

`r \to rate`

Solving (a): The expression for g

Since the rate is constant, the distribution of G follows:

`G(t) = G_0 * e^(rt)`

`G(1) = 120` implies that:

`G(t) = G_0 * e^(rt)`

`120 = G_0 * e^(r*1)`

Substitute
`G_0 = 100`

`120 = 100 * e^{r`

Divide both sides by 100

`1.2 = e^{r`

Take natural logarithm of both sides

`\ln(1.2) = \ln(e^r)`

`0.1823 = r`

`r = 0.1823`

So, the expression for G is:

`G(t) = G_0 * e^(rt)`

`G(t) = 100 *e^(0.1823t)`

Solving (b): t when G(t) = 300

We have:

`G(t) = 100 *e^(0.1823t)`

`300 = 100 *e^(0.1823t)`

Divide both sides by 100

`3 = e^(0.1823t)`

Take natural logarithm

`\ln(3) = \ln(e^(0.1823t))`

`1.099 = 0.1823t`

Solve for t

`t = (1.099)/(0.1823)`

`t = 6` --- approximated

Solving (c): When there will be no grass

Reduction at a rate of 80 tons per year implies that:

`G(t) = 100 *e^(0.1823t)- 80t`

To solve for t, we set G(t) = 0

`0 = 100 *e^(0.1823t)- 80t\\`

Rewrite as

`80t = 100 *e^(0.1823t)`

Divide both sides by 100

`0.8t = e^(0.1823t)`

Take natural logarithm of both sides

`\ln( 0.8t) = \ln(e^(0.1823t))`

`\ln( 0.8t) = 0.1823t`

Plot the graph of:
`\ln( 0.8t) = 0.1823t`

`t = 1.7`