Created on 2024-06-20Asked by Owen Scott (Solvelet student)
A ladder 10 feet long is leaning against a wall. If the bottom of the ladder slides away from the wall at a rate of 2 feet per second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?
Solution
To find how fast the top of the ladder is sliding down the wall: 1. **Set up the relationship using Pythagoras' theorem:** x2+y2=102, where x is the distance of the bottom of the ladder from the wall and y is the height of the ladder on the wall. 2. **Differentiate both sides with respect to time t:** 2xdtdx+2ydtdy=0. 3. **Plug in the known values:** x=6,dtdx=2 ft/s. Find y when x=6: y2=102−62=100−36=64⇒y=8. 4. **Substitute into the differentiated equation:** 2(6)(2)+2(8)dtdy=0,24+16dtdy=0⇒16dtdy=−24⇒dtdy=−1624=−1.5 ft/s. Therefore, the top of the ladder is sliding down the wall at a rate of 1.5 feet per second. Solved on Solvelet with Basic AI Model
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DefinitionRelated rates involves knowing how to derive one rate which is changing with another rate which is also changing (deriving) It is used many a times in problems where you need to track how two or more variables are changing with time. Ex: If a balloon radius grows at a 3 cm/s, how fast is the volume then expanding? Use dtdV=4πr2dtdr.
