The weight w of an astronaut (in pounds) is related to her height h above the surface of the earth (in miles) by w=w_0(3960/(3960+h))^2.  where 
w0
 is the weight of the astronaut on the surface of the earth. If the astronaut weighs 125 pounds on earth and is in a rocket, being propelled upward at a speed of 11 mi/s, find the rate at which her weight is changing (in lb/s) when she is 40 miles above the earth's surface. (Round your answer to four decimal places.)

The given equation for the astronaut's weight \( w \) as a function of height \( h \) above the Earth's surface is:

\[
w = w_0 \left( \frac{3960}{3960 + h} \right)^2
\]

where:
- \( w_0 = 125 \) pounds (the astronaut's weight on Earth),
- \( h \) is the height above the Earth's surface (in miles),
- \( \frac{dh}{dt} = 11 \) mi/s (the rate of change of height, i.e., velocity of the rocket).

We are asked to find the rate at which the astronaut's weight is changing (in lb/s) when she is 40 miles above the Earth's surface (i.e., when \( h = 40 \)).

### Step 1: Differentiate the equation for weight with respect to time.

The rate of change of weight with respect to time is \( \frac{dw}{dt} \), and we can apply the chain rule:

\[
\frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt}
\]

We know \( \frac{dh}{dt} = 11 \) mi/s, so now we need to compute \( \frac{dw}{dh} \).

### Step 2: Differentiate \( w = w_0 \left( \frac{3960}{3960 + h} \right)^2 \) with respect to \( h \).

Using the chain rule and power rule:

\[
\frac{dw}{dh} = w_0 \cdot 2 \left( \frac{3960}{3960 + h} \right) \cdot \left( -\frac{3960}{(3960 + h)^2} \right)
\]

Simplifying:

\[
\frac{dw}{dh} = -2w_0 \cdot \frac{3960^2}{(3960 + h)^3}
\]

### Step 3: Plug in the known values.

- \( w_0 = 125 \),
- \( h = 40 \).

Substitute these into the equation for \( \frac{dw}{dh} \):

\[
\frac{dw}{dh} = -2 \cdot 125 \cdot \frac{3960^2}{(3960 + 40)^3} = -250 \cdot \frac{3960^2}{4000^3}
\]

Now, compute \( \frac{dw}{dh} \):

\[
\frac{dw}{dh} = -250 \cdot \frac{15681600}{64000000000} \approx -0.06125
\]

### Step 4: Compute \( \frac{dw}{dt} \).

Using \( \frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt} \):

\[
\frac{dw}{dt} \approx -0.06125 \cdot 11 \approx -0.6738 \, \text{lb/s}
\]

Thus, the rate at which the astronaut's weight is changing when she is 40 miles above the Earth's surface is approximately \( -0.6738 \) lb/s.

### Final Answer:
\[
\boxed{-0.6738 \, \text{lb/s}}
\]

Would you like further clarification on any part of this process?

### Follow-up Questions:
1. How would the rate of change in weight differ if the astronaut were 100 miles above the surface?
2. What happens to the astronaut's weight change rate as height increases to infinity?
3. How would the problem change if the rocket's speed were doubled?
4. Can you derive a general formula for the rate of weight change \( \frac{dw}{dt} \) at any height \( h \)?
5. How would you interpret the negative sign in the weight change rate physically?

### Tip:
Always remember to check units in physics problems to ensure consistency, especially when mixing distances, times, and velocities!

The weight w of an astronaut (in pounds) is related to her height h above the surface of the earth (in miles) by w=w_0(3960/(3960+h))^2, where w0 is the weight of the astronaut on the surface of the earth. If the astronaut weighs 125 pounds on earth and is in a rocket, being propelled upward at a speed of 11 mi/s, find the rate at which her weight is changing (in lb/s) when she is 40 miles above the earth's surface. (Round your answer to four decimal places.)

Learn how to calculate the rate of change of an astronaut's weight when propelled at 11 mi/s and 40 miles above the Earth's surface using calculus and the chain rule. The astronaut weighs 125 pounds on Earth, and we determine how fast this changes at higher altitudes.

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