related rates: A weight is attached to one end of a rope....

[sickplaya]

A weight is attached to one end of a rope, of length 8.5m, passing over a pulley 4.5m above a horizontal floor. The other end of the rope is attached to a point, 0.5m above the floor, on the rear of a moving tractor. If the tractor is moving at a constant speed of 1m/sec, how fast is the weight rising when the tractor is 3m from the point directly below pulley?

i got an answer of 0.555 m/sec

 

[soroban]

Re: related rates

Hello, sickplaya!

A weight is attached to one end of a rope, of length 8.5m, passing over a pulley
4.5m above a horizontal floor. .The other end of the rope is attached to a point,
0.5m above the floor, on the rear of a moving tractor. .If the tractor is moving
at a constant speed of 1m/sec, how fast is the weight rising when the tractor
is 3m from the point directly below pulley?

                              P
                              *
                           *  |
                        *     |
                 y+4 *        |4.5-y
                  *           |
               *              |
            *                 * W 
         *                    |
    T *                       |
      |                       |y
   0.5|                       |
      |                       |
    - * - - - - - - - - - - - + -
      A                       B

The weight is \(\displaystyle W\), the pulley is \(\displaystyle P\), the truck is \(\displaystyle T.\)

The height of the weight is: \(\displaystyle y \,=\,WB.\)
Then: \(\displaystyle PW\,=\,4.5\,-\,y\,\) and: \(\displaystyle \,TP\:=\:8.5\,-\,(4.5\,-\,y)\:=\:y\,+\,4\)


At any time \(\displaystyle t\), the diagram looks like this:

                              P
                              *
                           *  |
                        *     |
                 y+4 *        |
                  *           |4
               *              |
            *                 | 
         *                    |
    T * - - - - - - - - - - - * Q
                  x

Let \(\displaystyle x\,=\,TQ.\)

Pythagorus tells us: \(\displaystyle \y\,+\,4)^2\:=\:x^2\,+\,4^2\;\) [1]

Differentiate with respect to time: \(\displaystyle \:2(y\,+\,4)\cdot\frac{dy}{dt}\:=\:2x\cdot\frac{dx}{dt}\;\) [2]


When \(\displaystyle x\,=\,3\), from [1], we have: \(\displaystyle \y\,+\,4)^2\:=\:3^2\,+\,16\;\;\Rightarrow\;\;y\,=\,1\)
We are told that: \(\displaystyle \,\frac{dx}{dt}\,=\,1\)

Substitute into [2]: \(\displaystyle \:2(5)\cdot\frac{dy}{dt}\:=\2)(3)(1)\)

Therefore: \(\displaystyle \L\:\fbox{\frac{dy}{dt}\:=\:0.6\text{ m/sec}}\)

 

[sickplaya]

yeah i made a really stupid mistake and i get your answer of 0.6 now. thanks a lot =)