A 15-ft ladder rests against a vertical wall. if the top of the ladder slides down the wall at a rate of 0.33 ft/sec, how fast, in ft/sec, is the bottom of the ladder sliding away from the wall, at the instant when the bottom of the ladder is 9 ft from the wall? answer with 2 decimal places. type your answer in the space below. if your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

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aristocles aristocles · Answer 1 · 2018-08-19T04:10:47+02:00

Let the corner of the wall and floor is our origin

Now the end of the ladder is on the floor is at distance x and top end on the wall is at distance y from this corner.

So we will use Pythagoras Theorem to find the length of the ladder.

[tex]x^2 + y^2 = L^2[/tex]

now if we differentiate whole equation with time

[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0[/tex]

now the speed of the end of the ladder at the end of the floor will be given as

[tex]\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}[/tex]

here [tex]\frac{dy}{dt}[/tex] is the speed of end of the ladder on the wall.

so here we also know that

[tex] 9^2 + y^2 = 15^2[/tex]

[tex]y = 12 ft[/tex]

now we will plug in all values in the equation

[tex]\frac{dx}{dt} = \frac{12}{9}*(0.33)ft/s[/tex]

[tex]v = 0.44 ft/s[/tex]

so the end of the ladder on the floor will move with speed v = 0.44 ft/s