Rate of change: find rate at which boat is approaching wharf

[Guest]

A boat is being winched towards a wharf by a rope which is attached to its bow, B. the winch, W, is 3m above B. the rope is wound in at 12m/min. (i.e. dr/dt = -12). Find the rate at which the boat is approaching the wharf when r = 5.

thanks in advance

 

[soroban]

Re: Rate of change question...

Hello, americo74!

Did you make a sketch?

A boat is being winched towards a wharf by a rope which is attached to its bow $\displaystyle B.$
The winch $\displaystyle W$ is 3 m above $\displaystyle B.$
The rope is wound in at 12 m/min. (i.e. $\displaystyle \frac{dr}{dt}\,=\,-12$).

Find the rate at which the boat is approaching the wharf when $\displaystyle r\,=\,5.$

                              * W
                          *   |
                r     *       |
                  *           | 3
              *               |
          *                   |
    B * - - - - - - - - - - - +
                  x

We have: $\displaystyle \,x^2\,+\,3^2\:=\:r^2$

Differentiate with respect to time: $\displaystyle \,2x\left(\frac{dx}{dt}\right)\:=\:2r\left(\frac{dr}{dt}\right)\;\;\Rightarrow\;\;\frac{dx}{dt}\:=\:\frac{r}{x}\left(\frac{dr}{dt}\right)$

We are given: $\displaystyle \,\frac{dr}{dt}\,=\,-12$
When $\displaystyle r\,=\,5$, we have: $\displaystyle \,x^2\,+\,3^2\:=\:5^2\;\;\Rightarrow\;\;x\,=\,4$

Therefore: \(\displaystyle \:\frac{dx}{dt}\:=\:\frac{5}{4}(-12)\:=\:\L-15\) m/min.

 

[Guest]

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