If the temperature is constant, then the atmospheric pressure P (in pounds/square inch) varies with the altitude above sea level h in accordance with the law where is the atmospheric pressure at sea level and k is a constant. If the atmospheric pressure is at sea level and at 5,000 ft, how fast is the atmospheric pressure changing with respect to altitude at an altitude of 15,000 ft? Give the answer to four decimal places.

Question

If the temperature is constant, then the atmospheric pressure P (in pounds/square inch) varies with the altitude above sea level h in accordance with the law  where  is the atmospheric pressure at sea level and k is a constant. If the atmospheric pressure is  at sea level and  at 5,000 ft, how fast is the atmospheric pressure changing with respect to altitude at an altitude of 15,000 ft? Give the answer to four decimal places.

Quizplus · Accepted Answer

Using the given law, we have:
P(h) = P0 * e^(-kh)
where P0 is the pressure at sea level and k is a constant.
Taking the derivative, we get:
P'(h) = -k * P0 * e^(-kh)

At an altitude of 15,000 ft, h = 15,000 ft. We are asked to find P'(15,000):
P'(15,000) = -k * P0 * e^(-k * 15,000)

To find the constant k, we can use the given information about pressure at sea level and at 5,000 ft:
P(0) = P0 = 11eb12b0_a99c_7817_b394_f1a4ab340c61_TB7866_11
P(5,000) = P0 * e^(-5k) = 11eb12b0_a99c_7818_b394_59e1c81ef983_TB7866_11

Dividing the second equation by the first, we get:
e^(-5k) = 11eb12b0_a99c_7818_b394_59e1c81ef983_TB7866_11 / 11eb12b0_a99c_7817_b394_f1a4ab340c61_TB7866_11 = 0.614
Taking the natural logarithm of both sides, we get:
-5k = ln(0.614)
k = -0.2813

Now substituting k and P0 into the expression for P'(15,000), we get:
P'(15,000) = -(-0.2813) * 11eb12b0_a99c_7817_b394_f1a4ab340c61_TB7866_11 * e^(-0.2813 * 15,000) = 11eb12b0_a99c_9f2a_b394_c9e0b42c5ee6_TB7866_11

Therefore, the answer is B.

If the Temperature Is Constant, Then the Atmospheric Pressure P