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JA6.

Statement

For any problem that you couldn’t solve at first, suggest a strategy you might try to tackle the problem and show what happened as a result.

  • Describe any strategic gaps you were unable to bridge
  • list three helpful insights that may help another person trying to tackle the problem

Solution

From the written assignment, I struggled a lot solving this problem:

An airplane is flying towards a radar station at a constant height of 6 km above the ground. If the distance s between the airplane and the radar station is decreasing at a rate of 400 km per hour when s=10 km., what is the horizontal speed of the plane? Make sure your answer includes units (UoPeople, 2023).

The problem is clearly a related rates problem, that involves changes in height, distance, speed, and time. The speed of the plane can be derived from the distance and time, however, the height can not be ignored as it makes the distance a function to the time as well.

My strategy to solve the problem was to look for a mathematical relationship between all variables, which I found to be the Pythagorean theorem: s^2 = h^2 + x^2

Then, I took the derivative of both sides with respect to time, and I got:

\[ \begin{aligned} 2s\frac{ds}{dt} &= 2h\frac{dh}{dt} + 2x\frac{dx}{dt} \\ s\frac{ds}{dt} &= h\frac{dh}{dt} + x\frac{dx}{dt} \\ \end{aligned} \]

Then, I substituted the values given in the problem, where s = 10, and ds/dt = -400, and we find x according to the Pythagorean theorem:

\[ \begin{aligned} x &= \sqrt{s^2 - h^2} \\ &= \sqrt{10^2 - 6^2} \\ &= \sqrt{64} \\ &= 8 \text{km} \\ \end{aligned} \]
\[ (10)(-400) = (6)\frac{dh}{dt} + (8)\frac{dx}{dt} \]

My problem was the rate of height change dh/dt is not given, so I treated it as a variable, and when I tried to solve the problem, I got an equation above with two variables: distance (x) and height change (dh/dt). I tried to solve it, but I couldn’t.

After looking on a few more problems and searching the internet I had a second look at the problem statement and noticed that the height is constant, and therefore dh/dt = 0, which I then plugged into the equation and solved for dx/dt:

\[ \begin{aligned} (10)(-400) &= (6)(0) + (8)\frac{dx}{dt} \\ -4000 &= 8\frac{dx}{dt} \\ \frac{dx}{dt} &= -500 \text{km/hr} \\ \end{aligned} \]

Thus, the horizontal speed of the plane is 500 km/hr.

I had a hard time solving this problem, thus I have a few notes that may help anyone tying to solve such problems:

  • Although the real world equivalent of the problem seems complicated and involves many more factor than mentioned in the problem, usually problem statements are simplified to make them easier to solve. Thus, it is important to read the problem carefully and make sure you understand what is given and what is not.
  • Take extra care for every word in the problem statement, as they may be important. For example, in this problem, the word “constant” was the key to solving the problem.
  • When you stuck, go back and read the problem statement again and again, rather than looking on random places; you may find something you missed.

References

  • UoPeople (2023). Math 1211: Calculus I. Unit 6: written assignment.