a ladder 20 ft. long rests against a vertical wall.If the bottom of a ladder slides away from the wall at a rate of 4 ft./sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 12 ft. from the wall?
x^2 +y^2 = 20^2
x^2 + y^2 = 400 Pythagoras Theorem
differentiate w.r.t to t
2x dx/dt + 2y dy/dt = 0
x dx/dt = -y dy/dt
now when x = 12, dx/dt = -4 (sliding away from the wall)
y = sqrt (400 - 144) = 16 ft
insert these in the differential eq. above to get dy/dt
(12)(-4) = -16dy/dt
dy/dt = 3 ft/sec
According to your answer of a positive dy/dt, the laddder would be sliding UP the wall.
Since the ladder is sliding away from the wall along he ground, dx/dt should be +4 ft/sec
which would make dy/dt = -3 ft/sec, showing that y is decreasing, thus sliding DOWN the wall as expected.
Your number-crunching is ok
To solve this problem, we can use the Pythagorean theorem. Let's call the distance between the bottom of the ladder and the wall "x" and the distance between the top of the ladder and the ground "y". According to the Pythagorean theorem, we have x^2 + y^2 = 20^2 (since the ladder is 20 feet long).
Now, let's differentiate both sides of this equation with respect to time, t. We get:
2x(dx/dt) + 2y(dy/dt) = 0
where dx/dt represents the rate at which the distance between the bottom of the ladder and the wall is changing, and dy/dt represents the rate at which the distance between the top of the ladder and the ground is changing.
Since the ladder is sliding away from the wall, dx/dt is -4 ft/sec.
Now, we need to find the value of x and y when the bottom of the ladder is 12 ft. from the wall. From the equation x^2 + y^2 = 20^2, we can solve for y:
y^2 = 20^2 - 12^2
y^2 = 400 - 144
y^2 = 256
y = sqrt(256)
y = 16 ft.
Now, we can substitute the values of x, dx/dt, y, and solve for dy/dt:
(12)(-4) + (16)(dy/dt) = 0
-48 + 16(dy/dt) = 0
16(dy/dt) = 48
dy/dt = 3 ft/sec.
Therefore, the top of the ladder is sliding down the wall at a rate of 3 ft/sec when the bottom of the ladder is 12 ft. from the wall.
To find how fast the top of the ladder is sliding down the wall, we can use the Pythagorean Theorem to relate the variables x and y. The equation is:
x^2 + y^2 = 20^2
Taking the derivative of both sides with respect to time t (differentiate w.r.t. t), we get:
2x dx/dt + 2y dy/dt = 0
Since the ladder is sliding away from the wall at a rate of 4 ft./sec (dx/dt = -4), we can substitute this into the equation:
x (-4) = -y dy/dt
When the bottom of the ladder is 12 ft. from the wall (x = 12), we can solve for dy/dt:
(12)(-4) = -16 dy/dt
dy/dt = 3 ft/sec
Therefore, the top of the ladder is sliding down the wall at a rate of 3 ft/sec.