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Asked by Himani Last Modified
Answer Let's denote:
According to the Pythagorean theorem, for any position of the ladder, we have x2+y2=52x2+y2=52, since the ladder is always 5 meters long.
Now, we differentiate both sides of this equation with respect to time tt, considering that xx is changing with time: ddt(x2+y2)=ddt(52)dtd(x2+y2)=dtd(52) 2xdxdt+2ydydt=02xdtdx+2ydtdy=0
Given that the ladder is being pulled away from the wall at a rate of dxdt=2dtdx=2 cm/s, and we want to find dydtdtdy when x=4x=4 m, we can plug in these values into the equation:
2(4)(2)+2ydydt=02(4)(2)+2ydtdy=0 16+4ydydt=016+4ydtdy=0 4ydydt=−164ydtdy=−16 dydt=−164ydtdy=−4y16
Now, we need to find yy when x=4x=4. Using the Pythagorean theorem: 42+y2=5242+y2=52 16+y2=2516+y2=25 y2=9y2=9 y=3y=3
Now, we can find dydtdtdy when y=3y=3: dydt=−164⋅3dtdy=−4⋅316 dydt=−43dtdy=−34
So, the height of the ladder on the wall is decreasing at a rate of 4334 meters per second when the foot of the ladder is 4 meters away from the wall.
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