# Rates of Change of a 15ft ladder

• Mar 21st 2009, 04:37 PM
ment2byours
Rates of Change of a 15ft ladder
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?
• Mar 21st 2009, 04:50 PM
Pinkk
$x^{2}+y^{2}=15^{2}$
Then differentiate with respect to $t$:

$2x\frac{dx}{dt}+ 2y\frac{dy}{dt} = 0$

So the rate of change we are interested in is when $x=9$, then $y=12$ (9,12,15 triangle). And $\frac{dx}{dt}=\frac{1}{2}$

Plug in the values for $x,y, \frac{dx}{dt}$ and solve for $\frac{dy}{dt}$

For the second part, $A=\frac{1}{2}xy$ so therefore $\frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2 }y\frac{dx}{dt}$ Plug in all your values and you will get your $\frac{dA}{dt}$ value.
• Mar 21st 2009, 05:33 PM
ment2byours
ok for a.) I got -3/8 and b.) 21/16

is that correct?
• Mar 21st 2009, 05:37 PM
Pinkk
Edit: Since your problem included units, I would be careful to include those units in your answers if you're ever taking a test. So $\frac{dy}{dt}=\frac{-3}{8} ft/sec$ and $\frac{dA}{dt}=\frac{21}{16} ft^{2}/sec$