+9665 1. A differentiation question that I never knew how to do:
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s = 400t - 16t2. How fast is the distance between the rocket and an observer on the ground 1800 feet away from the launching site changing when the rocket is still rising and is 2400 feet above the ground? (see Fig. 20-6)
I will include Fig. 20 - 6 here:
2. An integration question, I used trigonometrical substitution but got a different answer as the book's answer.
That looks easy though.
\(\text{Compute }\int\dfrac{dx}{x^2-9}\)
I got an answer of \(\dfrac{\ln|\dfrac{x-3}{\sqrt{x^2-9}}|}{9}+K,\text{where K = C/9}\) by substituting \(x = 3\sec \theta\). But the book's answer says it's \(\dfrac{1}{6}\ln|\dfrac{x-3}{x+3}|+C\). Why is that??
+33653 1. Here's a hint for part 1 Max.
I'm sure you can take it from here.
2. Here's part 2.
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+33653 Note: In part 1 my hint contains an error. \(\frac{du}{dt}=\frac{s}{\sqrt{s^2+1800^2}}\frac{ds}{dt}\) is the correct expression for the rate of change of u. The 2 shouldn't be there!
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