Length in cm for fish

Question

The length (in cm) of a particular species of fish at time t (in months) is given by:

L(t)= 34-32(k^t)

k a constant.

(a) If L (4) = 10, find the value of k for this particular species of fish.

Determinetherateatwhichthefishisgrowingatt=6months.

Answer 1

To find the value of k, we can substitute the given values into the equation and solve for k.

Given: L(t) = 34 - 32(k^t) and L(4) = 10

Substituting t = 4 and L(4) = 10 into the equation, we have:

10 = 34 - 32(k^4)

Rearranging the equation, we get:

32(k^4) = 34 - 10

32(k^4) = 24

Dividing both sides by 32, we get:

k^4 = 24/32

k^4 = 3/4

To find the value of k, we take the fourth root of both sides:

k = (3/4)^(1/4)

k ≈ 0.8706

So the value of k for this particular species of fish is approximately 0.8706.

Now, to determine the rate at which the fish is growing at t = 6 months, we can differentiate the length function L(t) with respect to t and evaluate it at t = 6.

L(t) = 34 - 32(k^t)

Differentiating both sides with respect to t, we get:

dL/dt = -32k^t * ln(k)

Substituting t = 6 and the value of k we found earlier, we have:

dL/dt = -32(0.8706^6) * ln(0.8706)

Calculating this expression will give us the rate at which the fish is growing at t = 6 months.

Please let me know if you need any further assistance!