Chapter 4, Problem 37RE

To determine

Tofind:how fast the distance between the boy and balloon increases at 3s later.

Answer to Problem 37RE

It increases at $\text{13ft/s}$ .

Explanation of Solution

Given:

  $\frac{\text{dh}}{\text{dt}}\text{=5ft/sec}$

  $\frac{\text{dx}}{\text{dt}}\text{=15ft/sec}$

Concept used:

According to the Pythagoras theorem:

  ${\text{s}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+h}}^{\text{2}}$ .

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{>0}\text{.}$ the concave will be open upward and local minima can be found.

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{<0}\text{.}$ the concave will open downward and local maxima can be found.

According to the Pythagoras theorem:

  ${\text{s}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+h}}^{\text{2}}$ .

Calculation:

Let the figure of the question be:

  

According to the given data: $\frac{\text{dh}}{\text{dt}}\text{=5ft/sec}$ and $\frac{\text{dx}}{\text{dt}}\text{=15ft/sec}$ .

According to the Pythagoras theorem:

  ${\text{s}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+h}}^{\text{2}}$ .

Differentiating with respect to t:

  $\text{2s}\frac{\text{ds}}{\text{dt}}\text{=2x}\frac{\text{dx}}{\text{dt}}\text{+2h}\frac{\text{dh}}{\text{dt}}$ .

  $\text{s}\frac{\text{ds}}{\text{dt}}\text{=x}\frac{\text{dx}}{\text{dt}}\text{+h}\frac{\text{dh}}{\text{dt}}$ .

  $\frac{\text{ds}}{\text{dt}}\text{=}\frac{\text{1}}{\text{s}}\left(\text{x}\frac{\text{dx}}{\text{dt}}\text{+h}\frac{\text{dh}}{\text{dt}}\right)$ .

To find the length at 3 seconds:

  $\text{x=15}\frac{\text{ft}}{\text{s}}\text{3s=45ft}$ .

  $\text{h=5}\frac{\text{ft}}{\text{s}}\text{.3s+45=60ft}$ .

  $\text{s=}\sqrt{{\text{45}}^{\text{2}}{\text{+60}}^{\text{2}}}\text{=}\sqrt{\text{5625}}\text{=75ft}$ .

Plug in known value for the moment in time:

  $\frac{\text{ds}}{\text{dt}}\text{=}\frac{\text{1}}{\text{s}}\left(\text{x}\frac{\text{dx}}{\text{dt}}\text{+h}\frac{\text{dh}}{\text{dt}}\right)$

  $\Rightarrow \frac{\text{ds}}{\text{dt}}\text{=}\frac{\text{1}}{75}\left(45\left(15\right)\text{+60}\left(5\right)\right)$ .

  $\Rightarrow \frac{975}{75}$

  $\frac{\text{ds}}{\text{dt}}\text{=13ft/s}$

Hence it increases at $\text{13ft/s}$ .