The atmospheric pressure P(h) (in pounds per square inch) at a height h (in miles) above sea level on Earth satisfies a differential equation P' = -kP for some positive constant k.

(a) Measurements with a barometer show that P(0) = 1.2 and P(9) = 0.7. What is the decay constant k?
(b) Determine the atmospheric pressure 14 miles above sea level.

P' = -kP

dP/P = -k dh
lnP = -kh + c

P = c e^-kh
P(0) = 1.2, so c = 1.2

P = 1.2 e^-kh
1.2 e^(-9k) = 0.7
k = .06

so,

P(h) = 1.2 e^-.06h

(a) Well, well, looks like we have a differential equation here. Don't you just love those? So, let's solve this puzzle, shall we?

First things first, we need to determine the decay constant k. We know that P' = -kP, but we also know that P(0) = 1.2. That's our starting point.

Now, let's integrate both sides of the equation with respect to h. We end up with ln(P) = -kh + C, where C is a constant of integration.

Next, let's determine the value of C. We know that P(0) = 1.2, so we plug that into our equation. We get ln(1.2) = -k(0) + C. Well, since k(0) = 0, C = ln(1.2).

Now, let's move on to Part B.

(b) To determine the atmospheric pressure 14 miles above sea level, we need to substitute the values we know into our equation. So, we have ln(P) = -kh + ln(1.2), and we know P(9) = 0.7.

Using P(9) = 0.7, we plug in the values h = 9 and P = 0.7 into our equation. We get ln(0.7) = -k(9) + ln(1.2).

Now, we can solve for k. Rearranging the equation, we have -k(9) = ln(0.7) - ln(1.2). Divide both sides by -9, and we get k = (ln(1.2) - ln(0.7))/9.

So, after all of that, the decay constant k is (ln(1.2) - ln(0.7))/9.

Now, remember folks, always keep your barometer close and your jokes even closer!

To find the decay constant k, we need to solve the differential equation P' = -kP using the given information.

(a) First, let's solve the differential equation using separation of variables. We can rewrite it as:

1/P dP = -k dh

Integrating both sides:

∫ 1/P dP = -∫ k dh

ln|P| = -kh + C

where C is the constant of integration.

Using the initial condition P(0) = 1.2, we can substitute h = 0 and P = 1.2:

ln|1.2| = -k(0) + C
ln|1.2| = C

Therefore, the equation becomes:

ln|P| = -kh + ln|1.2|

Next, using the second initial condition P(9) = 0.7:

ln|0.7| = -k(9) + ln|1.2|

Simplifying:

-ln|0.7| = -9k + ln|1.2|

ln|0.7| + ln|1.2| = 9k

ln|0.7 * 1.2| = 9k

ln|0.84| = 9k

Now, solving for k:

k = ln|0.84| / 9 ≈ -0.0289

So, the decay constant k is approximately -0.0289.

(b) To determine the atmospheric pressure at 14 miles above sea level, we can use the general solution we found in part (a) and substitute h = 14:

ln|P| = -k(14) + ln|1.2|

ln|P| = -0.0289(14) + ln|1.2|

ln|P| ≈ -0.4055 + ln|1.2|

ln|P| ≈ -0.4055 + ln(1.2)

Using logarithmic properties:

ln|P| ≈ ln(1.2/ e^0.4055)

ln|P| ≈ ln(1.2/ 0.6664)

Taking the exponential of both sides:

|P| ≈ 1.2/0.6664

Therefore, the atmospheric pressure 14 miles above sea level is approximately |P| ≈ 1.8 pounds per square inch.

To solve the differential equation, P' = -kP, we can use separation of variables. This involves isolating the variables P and h on separate sides of the equation. Let's go through the steps to find the value of the decay constant k and the atmospheric pressure at 14 miles above sea level.

(a) We are given that P(0) = 1.2 and P(9) = 0.7. We'll use this information to determine the value of the decay constant k.

First, let's separate the variables:

P' / P = -k

Next, integrate both sides of the equation with respect to h:

∫(P' / P) dh = ∫(-k) dh

Applying the integral rules, we get:

ln|P| = -kh + C

where C is the constant of integration. Now, let's solve for P in terms of h:

|P| = e^(-kh + C)

Since e^C is just a constant, we can combine it with the absolute value:

|P| = Ce^(-kh)

where C = e^C.

Now, let's evaluate the constant C using the given information:

When h = 0, P = 1.2
|1.2| = Ce^(0) as any number to the power of 0 is 1.
1.2 = C

Now our equation becomes:

|P| = 1.2e^(-kh)

Next, let's use the second set of data to find k:

When h = 9, P = 0.7
|0.7| = 1.2e^(-9k)
0.7 = 1.2e^(-9k)

Now, solve for k:

e^(-9k) = 0.7 / 1.2
e^(-9k) = 0.58333...

Take the natural logarithm ln of both sides:

-9k = ln(0.58333...)

Now, solve for k:

k = -ln(0.58333...) / 9

Using a calculator or software, we find:

k ≈ 0.036915

So, the decay constant k is approximately 0.036915.

(b) To determine the atmospheric pressure 14 miles above sea level, we can use the equation we derived earlier:

|P| = 1.2e^(-kh)

Substitute the value of k we found:

|P| = 1.2e^(-0.036915h)

Now plug in h = 14:

|P| = 1.2e^(-0.036915 * 14)

Using a calculator or software, we find:

|P| ≈ 0.562

Remember, we must take the absolute value because the atmospheric pressure can't be negative. Therefore, the atmospheric pressure 14 miles above sea level is approximately 0.562 pounds per square inch.