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Related Rates Promlem, Please Help!
A bird of prey is perched at the top of a tree that is 40 m high. The bird watches as a delectable Kingston squirrel runs away from the base of the tree at a rate of 2 m/s. What is the rate of change of the distance between the bird and the squirrel when the squirrel is 30 m from the tree?
I know:
y=height of tree so y=40 m
x=distance of base from tree so x= 30 m
d=distance from bird to squirrel
dx/dt=2m/s
dd/dt=? when squirrel is 3 m from tree
with pythag theorm: 40^2+30^2=50^2
so d= 50 m
x2 + y2 = d2
f'(x) = 2x (dx/dt) + 2y(dy/dt)=2d(dd/dt)
This is how far i have come and now i am stuck. Please help with this question. Thank you
Homework Statement
A bird of prey is perched at the top of a tree that is 40 m high. The bird watches as a delectable Kingston squirrel runs away from the base of the tree at a rate of 2 m/s. What is the rate of change of the distance between the bird and the squirrel when the squirrel is 30 m from the tree?
I know:
y=height of tree so y=40 m
x=distance of base from tree so x= 30 m
d=distance from bird to squirrel
dx/dt=2m/s
dd/dt=? when squirrel is 3 m from tree
with pythag theorm: 40^2+30^2=50^2
so d= 50 m
x2 + y2 = d2
f'(x) = 2x (dx/dt) + 2y(dy/dt)=2d(dd/dt)
This is how far i have come and now i am stuck. Please help with this question. Thank you