The size of a population of fish in a pond is modeled by the function P, where P(t) gives the number of fish and t gives the number of years after the first year of introduction of the fish to the pond for 0≤t≤10. The graph of the function P and the line tangent to P at t=4 are shown above. Which of the following gives the best estimate for the instantaneous rate of change of P at t=4 ?

At one moment this same instantaneous modification rate would be the same with the functionality derivative assessed somewhere at the time. In certain other terms, the pitch of such tangent line towards the curvature is somewhere at a level equivalent to that as well.

According to the question,

(x₁, y₁) = (3, 0)
(x₂, y₂) = (6, 838)

As we know,

⇒ [tex]Slope (m) = \frac{y_2-y_1}{x_2-x_1}[/tex]

By substituting the values, we get

⇒ [tex]=\frac{838-0}{6-3}[/tex]

⇒ [tex]=\frac{838}{3}[/tex]

⇒ [tex]=279.33[/tex]

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MrRoyal MrRoyal · Answer 2 · 2021-08-24T04:46:32+02:00

The instantaneous rate of change is the rate of change at a particular point or instant in the graph. The instantaneous rate of change of P at [tex]t=4[/tex] is 250

I've added as an attachment, the graph that represents the population of the fish.

To determine the instantaneous rate of change at [tex]t=4[/tex], we simply calculate the slope (m) of the tangent line using:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Because of the scale of the graph, I will use the following estimated points from the tangent line

[tex](x_1,y_1) = (3,0)[/tex]

[tex](x_2,y_2) = (4,250)[/tex]

So, the slope is:

[tex]m = \frac{250 - 0}{4 - 3}[/tex]

[tex]m = \frac{250}{1}[/tex]

[tex]m = 250[/tex]

Hence, the instantaneous rate of change of P at [tex]t=4[/tex] is 250

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