A ladder 10 feet long is leaning against a house. When the top of the ladder is 8 feet above the ground, it is sliding downward at the rate of 2 feet per second. How rapidly is the base of the ladder sliding away from the house?
I know you have to use pythagorean theorem, but I'm not getting the right answer.
a. 5/3
b. 2
c. 7/3
d. 8/3
e. 3
I know you have to use pythagorean theorem, but I'm not getting the right answer.
a. 5/3
b. 2
c. 7/3
d. 8/3
e. 3
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Let x = distance from house to base of ladder
Let y = distance from ground to top of ladder
By Pythagorean theorem, we get:
x² + y² = 10²
Integrating with respect to t (time), we get:
2x dx/dt + 2y dy/dt = 0
2x dx/dt = -2y dy/dt
dx/dt = -y/x dy/dt
Now we plug in known values:
Top of ladder is 8 ft above the ground: y = 8, x = √(10²-8²) = 6
Ladder is sliding downward at the rate of 2 ft/sec: dy/dt = -2
dx/dt = -8/6 * -2
Answer: d. 8/3
Mαthmφm
Let y = distance from ground to top of ladder
By Pythagorean theorem, we get:
x² + y² = 10²
Integrating with respect to t (time), we get:
2x dx/dt + 2y dy/dt = 0
2x dx/dt = -2y dy/dt
dx/dt = -y/x dy/dt
Now we plug in known values:
Top of ladder is 8 ft above the ground: y = 8, x = √(10²-8²) = 6
Ladder is sliding downward at the rate of 2 ft/sec: dy/dt = -2
dx/dt = -8/6 * -2
Answer: d. 8/3
Mαthmφm