Math Problem Statement

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.)

Solution

The given equation for the astronaut's weight ww as a function of height hh above the Earth's surface is:

w=w0(39603960+h)2w = w_0 \left( \frac{3960}{3960 + h} \right)^2

where:

  • w0=125w_0 = 125 pounds (the astronaut's weight on Earth),
  • hh is the height above the Earth's surface (in miles),
  • dhdt=11\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=40h = 40).

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

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

dwdt=dwdhdhdt\frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt}

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

Step 2: Differentiate w=w0(39603960+h)2w = w_0 \left( \frac{3960}{3960 + h} \right)^2 with respect to hh.

Using the chain rule and power rule:

dwdh=w02(39603960+h)(3960(3960+h)2)\frac{dw}{dh} = w_0 \cdot 2 \left( \frac{3960}{3960 + h} \right) \cdot \left( -\frac{3960}{(3960 + h)^2} \right)

Simplifying:

dwdh=2w039602(3960+h)3\frac{dw}{dh} = -2w_0 \cdot \frac{3960^2}{(3960 + h)^3}

Step 3: Plug in the known values.

  • w0=125w_0 = 125,
  • h=40h = 40.

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

dwdh=212539602(3960+40)3=2503960240003\frac{dw}{dh} = -2 \cdot 125 \cdot \frac{3960^2}{(3960 + 40)^3} = -250 \cdot \frac{3960^2}{4000^3}

Now, compute dwdh\frac{dw}{dh}:

dwdh=25015681600640000000000.06125\frac{dw}{dh} = -250 \cdot \frac{15681600}{64000000000} \approx -0.06125

Step 4: Compute dwdt\frac{dw}{dt}.

Using dwdt=dwdhdhdt\frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt}:

dwdt0.06125110.6738lb/s\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 0.6738-0.6738 lb/s.

Final Answer:

0.6738lb/s\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 dwdt\frac{dw}{dt} at any height hh?
  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!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Related Rates

Formulas

w = w_0 (3960/(3960 + h))^2
dw/dt = dw/dh * dh/dt

Theorems

Chain Rule of Differentiation

Suitable Grade Level

Grades 11-12 (Advanced Calculus)