Then differentiate with respect to :
So the rate of change we are interested in is when , then (9,12,15 triangle). And
Plug in the values for and solve for
For the second part, so therefore Plug in all your values and you will get your value.
A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall as shown in the figure. The bottom of the ladder (X) is moved away from the building at a constant rate of ½ foot per second.
a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building.
b) Find the rate of change in square feet per second of the area of triangle XOY when X is 9 feet from the building.
So I know I have to use the pythagorean theorem so I statred out with √(225-x(t)^2). Then I have truble finding its deriv. Can someoe help me and with (b) too? Do i use the general area equation A(t)=[x(t)*y(t)]2 first with that one?
Then differentiate with respect to :
So the rate of change we are interested in is when , then (9,12,15 triangle). And
Plug in the values for and solve for
For the second part, so therefore Plug in all your values and you will get your value.