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# A ladder 5m long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate 2cm/s. how fast is its height on the wall decreasing when the foot of the ladder is 4m away from the wall.

Let's denote: xx as the distance from the base of the ladder to the wall. yy as the height of the ladder on the wall. 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...

Let's denote:

• xx as the distance from the base of the ladder to the wall.
• yy as the height of the ladder on the wall.

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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