Question

Ocean sunfishes are well-known for rapidly gaining a lot of weight on a diet based on jellyfish.

The relationship between the elapsed time,
�
tt, in days, since an ocean sunfish is born, and its mass,
�
(
�
)
M(t)M, left parenthesis, t, right parenthesis, in milligrams is modeled by the following function:
�
(
�
)
=
4
⋅
(
81
49
)
�
M(t)=4⋅(
49
81
​
)
t
M, left parenthesis, t, right parenthesis, equals, 4, dot, left parenthesis, start fraction, 81, divided by, 49, end fraction, right parenthesis, start superscript, t, end superscript
Complete the following sentence about the rate of change in the mass of the sunfish. Round your answer to two decimal places.
The sunfish gains
2
7
7
2
​
start fraction, 2, divided by, 7, end fraction of its mass every
days.

Answer 1

Explanation:

We can use the formula for exponential growth to find the rate at which the sunfish gains mass:

M(t) = M(0) * e^(kt)

where:

M(0) = the initial mass of the sunfish (which we don't know)

k = the growth rate constant (which we also don't know yet)

t = the elapsed time since the sunfish was born (in days)

e = the mathematical constant approximately equal to 2.71828...

We are given the formula for M(t), which is:

M(t) = 4 * (81/49)^t

This means that M(0) = 4 * (81/49)^0 = 4.

To find the growth rate constant k, we can use the fact that the sunfish gains 2/7 of its mass every day. This means that the daily growth rate is 2/7 of the current mass. In other words, the daily growth rate is:

(2/7) * M(t)

We can relate this to the exponential growth formula by taking the derivative of M(t) with respect to t:

dM/dt = k * M(t)

Taking the derivative of M(t), we get:

dM/dt = ln(81/49) * 4 * (81/49)^t

Setting this equal to the daily growth rate, we get:

(2/7) * M(t) = ln(81/49) * 4 * (81/49)^t

Simplifying this expression, we get:

k = ln(81/49) * 4/7

Using a calculator, we can evaluate this expression to get:

k ≈ 0.0919

Therefore, the rate at which the sunfish gains mass is given by:

dM/dt = 0.0919 * M(t)

To find out how much mass the sunfish gains in one day, we can substitute M(t) = 4 * (81/49)^t into this formula and evaluate it at t = 1 (since we want to know the daily rate of change):

dM/dt = 0.0919 * 4 * (81/49)^1

dM/dt ≈ 0.54

This means that the sunfish gains 0.54 milligrams every day. To express this as a fraction of its mass, we can divide by the current mass:

(0.54/4) ≈ 0.1375

So the sunfish gains approximately 2/7 (or 0.2857) of its mass every 2.08 (or 7/2.54) days. Rounded to two decimal places, this is 0.14 or approximately 2/7 of its mass every 2.08 days.