A 11 - foot ladder is placed against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 3.2 ft/s. How fast is the top of the ladder sliding down the wall when the top of the ladder is 8 feet from the ground?
Solution: Let the y-axis represent the wall and let the x-axis represent the ground.
Let x=x(t) be the distance of the bottom of the ladder to the wall at time t and let y=y(t) be the distance of the top of the ladder to the ground at time t.
The question is to find the rate of change of y with respect to t, which is the same as to find the derivative dy /dt, if y=8 and dx /dt=3.2
Suppose the time at which the ladder is at this position is t=t1. By the Pythagorean theorem we get the equation x2+y2=("11")^2. This implies that if y=y(t1)="8" then x=x(t1)="7.549" . Therefore if we differentiate both sides of the aforementioned equation with respect to t then, for y=8, dx/dt=3.2, and x="7.55" , we get that dy/dt= "-2.831".
Hence, if the bottom of the ladder slides away from the wall at a constant rate of 3.2 ft/s then the top of the ladder slides down at the rate "2.831".
Where did I make mistakes? The numbers in " " need corrections.
Help me. Thanks
Solution: Let the y-axis represent the wall and let the x-axis represent the ground.
Let x=x(t) be the distance of the bottom of the ladder to the wall at time t and let y=y(t) be the distance of the top of the ladder to the ground at time t.
The question is to find the rate of change of y with respect to t, which is the same as to find the derivative dy /dt, if y=8 and dx /dt=3.2
Suppose the time at which the ladder is at this position is t=t1. By the Pythagorean theorem we get the equation x2+y2=("11")^2. This implies that if y=y(t1)="8" then x=x(t1)="7.549" . Therefore if we differentiate both sides of the aforementioned equation with respect to t then, for y=8, dx/dt=3.2, and x="7.55" , we get that dy/dt= "-2.831".
Hence, if the bottom of the ladder slides away from the wall at a constant rate of 3.2 ft/s then the top of the ladder slides down at the rate "2.831".
Where did I make mistakes? The numbers in " " need corrections.
Help me. Thanks
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x^2 + y^2 = 11^2 --------(1)
where x is horizontal distance, y is vertical distance when y = 8, x = √(121 - 64) = √57
differentiate eqn (1)
2x dx/dt + 2y dy/dt = 0
=> x dx/dt + y dy/dt = 0
√57 (3.2) + 8 dy/dt = 0
=> dy dt = - √57 (3.2) /8 = -√57 (0.4) = - 3.0199 ft/s
where x is horizontal distance, y is vertical distance when y = 8, x = √(121 - 64) = √57
differentiate eqn (1)
2x dx/dt + 2y dy/dt = 0
=> x dx/dt + y dy/dt = 0
√57 (3.2) + 8 dy/dt = 0
=> dy dt = - √57 (3.2) /8 = -√57 (0.4) = - 3.0199 ft/s