QA013

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course Mth 173

8/2 3:43

013. Applications of the Chain Rule*********************************************

Question: `qNote that there are 4 questions in this assignment.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q001. The Fahrenheit temperature T of a potato just taken from the oven is given by the function T(t) = 70 + 120 e^(-.1 t), where t is the time in minutes since the potato was removed from the oven. At what rate is the temperature changing at t = 5?

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Your solution:

T(5) = 70 + 120 e^(-.1(5))

= 70 + 72.78

= 142.78

142.78 degrees / 5 sec = 28.56 degrees/sec

confidence rating #$&*: Confused

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Given Solution:

`aThe rate of temperature change is given by the derivative function ( T ( t ) ) ', also written T ' (t).

Since T(t) is the sum of the constant function 70, whose derivative is zero, and 120 times the composite function e^(-.1 t), whose derivative is -.1 e^(-.1 t), we see that T ' (t) = 120 * ( -.1 e^(-.1 t) ) = -12 e^(-.1 t).

Note that e^(-.1 t) is the composite of f(z) = e^z and g(t) = -.1 t, and that its derivative is therefore found using the chain rule.

When t = 5, we have T ' (5) = -12 e^(-.1 * 5 ) = -12 e^-.5 = -7.3, approx.. This represents rate = change in T / change in t in units of degrees / minute, so at t = 5 minutes the temperature is changing by -7.3 degrees/minute.

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Self-critique (if necessary):

I did not know that I was supposed to get the derivative of the function first to find out the average rate at which the temperature changes.

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Self-critique Rating:

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The question asked for the rate, not the average rate.

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Question: `q002. The weight in grams of a growing plant is closely modeled by the function W(t) = .01 e^(.3 t ), where t is the number of days since the seed germinated. At what rate is the weight of the plant changing when t = 10?

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Your solution:

f(z) = e^z f’(z) = e^z

g(x) = .3t g’(x) = .3

y’ = g’ * f’(g(x))

y’ = .01 * .3 e^(.3 t)

y’ = .003 e^(.3t)

y’(10) = .003 e^(.3(10))

= .06

confidence rating #$&*: A little more confident

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Given Solution:

`aThe rate of change of weight is given by the derivative function ( W ( t ) ) ', also written W ' (t).

Since W(t) is .01 times the composite function e^(.3 t), whose derivative is .3 e^(.3 t), we see that W ' (t) = .01 * ( .3 e^(.3 t) ) = .003 e^(.3 t).

Note that e^(.3 t) is the composite of f(z) = e^z and g(t) = .3 t, and that its derivative is therefore found using the chain rule.

When t = 10 we have W ' (10) = .003 e^(.3 * 10) = .03 e^(3) = .06. Since W is given in grams and t in days, W ' will represent change in weight / change in clock time, measured in grams / day.

Thus at t = 10 days the weight is changing by .06 grams / day.

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Self-critique (if necessary):

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Self-critique Rating: