Aladder 10 ft long rests against a vertical wall. if the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (that is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)

The angle's rate of change when the bottom of the ladder is 8ft from the wall is

It is given the Length of Ladder is . and the distance between the bottom of the ladder to the wall is as shown in the below figure.

By using the Pythagoras Theorem

and

Differentiating both sides with respect to by using the chain rule

given

putting this in eq.1, we get

So the angle's rate of change when the bottom of the ladder is 8ft from the wall is.

Know more about Pythagoras Theorem here:

link

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."