A dog chases a squirrel at a speed of 12 m/s. The squirrel dashes up a tree trunk at a rate of 6 m/s. Find the rate of change of distance between the squirrel and the dog at an instant when the do is 12m from the tree trunk and the squirrel is 5m up the trunk.

Question

sriram · Answer

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sriram · Answer

bad drawin but cosider the vertical line as tree
horizontal as land on which d dog is runnin

anonymous · Answer

I have the answer and everything but I just don't get why you have to put 12 m/s in for dy/dt further into the question..

sriram · Answer

now let x be dist of dog from tree
and y dist of squ. from land
dist betwn them is L
L^2 = x^2 + y^2

anonymous · Answer

Yeah you take the derivative of the Pythagorean theorem and then plug in 12 m/s for dy/dt ? What does dy/dt even stand for?

sriram · Answer

differentiate wrt time
2L.dL/dt= 2x.dx/dt +2y.dy/dt
we know x=12 y=5 givin L=13
and dx/dt=12 and dy/dt=6
find dL/dt

anonymous · Answer

Ok thankss

sriram · Answer

dy/dt is rate of change of y
which is the squirel's speed