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A balloon is rising vertically above a level, straight road at a constant rate of $1$ ft/sec. Just when the balloon is $65$ ft above the ground, a bicycle moving at a constant rate of $ 17$ ft/sec passes under it. How fast is the distance between the bicycle and the balloon is increasing $3$ seconds later?

I am at a loss what to begin with? just a hint would do..

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  1. Find the height of the balloon at time $t$ seconds

  2. Find the horizontal distance of the bicycle at time $t$ seconds

  3. Find the distance between the balloon and the bicycle at time $t$ seconds

  4. Differentiate with respect to $t$

  5. Let $t = 3$.

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  • $\begingroup$ Using $s^2=x^2+y^2$ and differentiating both sides w.r.t $t$ ,substitute the values of $dx/dt$ and$ dy/dt$ and put $t=3$ Is that so? $\endgroup$ – Tom Lynd Oct 15 '13 at 14:29
  • $\begingroup$ I would differentiate amWhy's expression $\endgroup$ – Henry Oct 15 '13 at 15:18
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Let $v(t)$ be the height (vertical distance} of the balloon at time $t$, where $t$ is measured in seconds. Then $$v(t) = 65 + (1\text{ ft./sec}) \;\cdot t \;\text{sec} = 65 + t \;\text{ft.}.$$

Let $h(t)$ be the horizontal distance of the bike at time $t$, where $t$ is measured in seconds. Then $$h(t) = (17 \;\text{ft/sec}) \cdot t \;\text{sec}\; = 17 t \;\text{ft.}$$

The distance between the balloon and the bicycle at time $t$ seconds is the hypotenuse formed of the triangle formed by the vertical distance and horizontal distance, as legs: $$d(t) = \sqrt{(h(t))^2 + (v(t))^2} = \sqrt{(65 + t)^2 + (17t)^2}$$

Find $d'(t)$ and once you've done that, evaluate at $t = 3\;\text{sec.}$

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  • $\begingroup$ Nice write - up +1 $\endgroup$ – Amzoti Oct 16 '13 at 0:42

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