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Question
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.
According to Pythagoras : y(t)=√[15^2-x(t)^2]=√[225-x(t)^2].
So y'(t)=[2x(t)*x'(t)]/2√[225-x(t)^2]=x(t)/√[225-x(t)^2].
y'(x=9)=9/√(225-81)=3/4 feet/sec.
The area A(t)=x(t)*y(t) So A'(t)=x'(t)y(t)+x(t)y'(t)
A'(x=9)=(1/2)y(9)+9*(3/4). Now let's calculate y(9) :
y(9)=√[225-81]=12. Hence, A'(x=9)=(1/2)*12+9*(3/4)=12.75 square feet per second .
Alon.
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Alon Mandes
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