my teachers key is riddled with issues and this is our test review. A person flying a kite holds the string 5 feet above the ground level and the string is let out at a rate of 2 ft/sec as the kite moves horizontally at an altitude of 105 feet. assuming there is no sag the string, find the rate at which the kite is. moving when the string is 125 feet long.

  • $\begingroup$ What attempts have you made? $\endgroup$ – Harry Alli Nov 27 '17 at 6:42
  • $\begingroup$ ok so so far I used pythag. to get 75 for the length of the string. and I got an answer of 3.3 ft/sec but I'm not sure if that is correct $\endgroup$ – Brandon Foos Nov 27 '17 at 6:44

Let $s$ be the horizontal distance of the kite in feet at time $t$ in seconds. The length of the string is $2t$.

Pythagoras gives: $$(2t)^2=s^2+(100)^2$$

The easiest approach now is to use implicit differentiation with respect to $t$ to give $$8t=2s\frac {ds}{dt}$$ or $$\frac {ds}{dt}=\frac {4t}s$$

The length of the string at the point you are interested in is $2t=125$, and pythagoras gives $s=75$ (note this is not the length of the string but the horizontal distance in feet. We therefore have $$\frac {ds}{dt}=\frac {2\times 125}{75}=\frac {10}3 \text { feet per second}$$And this accords with the answer you put in your comment.

Note: implicit differentiation uses the chain rule $$\frac {d (f(y))}{dx}=\frac {d(f(y))}{dy}\cdot\frac {dy}{dx}$$

Here it is simpler than taking square roots to find an explicit equation for $s$ in terms of $t$.


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