J. Coroneos more rates of change q (1 Viewer)

sooki0326 · May 29, 2010

A weight W is attached to a rope 15m long which passes over a pully 8m above the ground.The other end of the rope is attached to a truck at a point 2m above the ground. If the truck moves off at a speed of 2.5 m/s, find how fast the weight is rising when it is 3m above the ground. (Hint: At time t, let the truck be x m from the vertical below the weight, and let the weight be y m above the ground. The show that (7+y)^2 = X^2 + 6^2.

I really dont have any idea as to how this looks. Help please~
P.S. The answer at the back says 2m/s

twistedrebel · May 29, 2010

lol i have done this question, 2ish weeks ago. I ll upload it later this evening/first thing in the morning if no one does it by then

Rezen · May 31, 2010

From the diagram, the rope is 15 m long.
i.e 8-y+hypotenuse=15
therefore, hypotenuse=y+7
and by pythagorus, x^2 +6^2 =(y+7)^2

When y=3, x=8. Also it is given that dx/dt=2.5m/s

$x^2+6^2=(y+7)^2 \\ \textup{implicitly differentiating},2x=2(y+7)\frac{\mathrm{d} y}{\mathrm{d} x} \\ \therefore \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x}{(y+7)}, at\; x=8, \; y=3, \; \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{8}{10} \\ \frac{\mathrm{d} y}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} x}\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{8}{10}.2.5=2m/s$

J. Coroneos more rates of change q (1 Viewer)

sooki0326

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twistedrebel

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Rezen

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