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Anni [7]
4 years ago
14

AP CALC HELP!!! Worth 38 points. AP Calc AB differental FRQ

Mathematics
1 answer:
sveta [45]4 years ago
5 0

Answer:

5a.  approximately 6 grams remain after 2 seconds

5b.  The graph shown cannot be a solution. The solution has negative slope everywhere.

5c.  y = 50/(t+5)

5d.  The amount is changing at a decreasing rate. (As y gets smaller, so does the magnitude of dy/dt.)

Step-by-step explanation:

5a. The tangent line has the equation ...

  y = f'(0)t +f(0)

Here, that is

  y = -0.02·10²·t +10 = 10 -2t

Then at t=2, the value is ...

  y = 10 -2·2 = 6 . . . . grams remaining

__

5b. y² is always positive (or zero), so -0.02y² will be negative. This is dy/dt, the slope of the curve with respect to time, so any curve with positive slope somewhere cannot be a solution.

__

5c. The equation is separable so can be solved by integrating ...

  ∫y^-2·dy = -0.02∫dt

  -y^-1 = -0.02t +c . . . . for some arbitrary constant c

Multiplying by -50 gives ...

  50/y = t + c . . . . for some constant c

We can find the value of c by invoking the initial condition. At t=0, y=10, so we have ...

  50/10 = 0 +c = 5

Then, solving for y, we get ...

  y = 50/(t+5)

__

5d. As noted above (and as described by the differential equation), the magnitude of the rate of change is proportional to the square of y. As y decreases, its rate of change will also decrease (faster). You can see that the curve for y flattens out as t increases. The amount of the substance is changing at a decreasing rate.

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