A cylindrical tank, 6 foot in radius, lies on it's side parallel and against the side of a warehouse. A ladder leans against the building, passes over and just touches the tank, and has a slope of -3/4. Find the equation of the ladder and the length of the ladder

Answer 1

use the equation y=mx+b

Answer 2

Put the diagram on the x-y grid
so the centre of the circle for the cylinder is C(6,6)
Let P(x,y) be the point of contact of the ladder with the cylinder.
Slope of ladder touching at P is -3/4
So the slope of PC = 4/3

(y-6)/(x-6) = 4/3
4(y-6) = 3(x-6)
y-6 = (3/4)(x-6) #1

equation of circle is
(x-6)^2 + (y-6)^2 = 36
but P lies on this, so
(x-6)^2 + (9/16)(x-6)^2 = 36 #2
times 16
16(x-6)^2 + 9(x-6)^2 = 576
25(x-6)^2 = 576
take √
5(x-6) = 24
x-6 = 24/5
x = 24/5 + 6 = 10.8
subbing back into #1, we get
y = 9.6

so P is (10.8 / 9.6)

equation for ladder:
let y = (-3/4)x + b
9.6 = (-3/4)(10.8) = b
b = 17.7
equation is y = -.75x + 17.7

x - intercept is 23.6
y - intercept is 17.7

length of ladder L
L^2 = 23.6^2 + 17.7^2 = 870.25
L = 29.5

Answer 3

near the 2/3 mark of solution should say

so P is (10.8 , 9.6)

typo had no effect on rest of solution

Answer 4

I think you get wrong at the slope of equation #1, and it makes your answer after #1 is wrong also.. but I think your procedure is correct.. Thank You

Answer 5

@Reiny I think you get wrong at the slope of equation #1, and it makes your answer after #1 is wrong also.. but I think your procedure is correct.. Thank You