Question

A cylindrical tank, 6 ft in radius, lies on its side parallel to and against the side of a warehouse. A ladder, which lean against the wall of the building, passes over and just touches the tank, and has a slope of – ¾. How long is the ladder?

Answer 1

Draw a picture. If the ladder reaches up y ft on the wall, and touches the ground x ft from the wall, then
(y-6)/(x-8) = 3/4
(y-6)/12 = 3/4
Looks like x-8=12
y-6 = 9
so the ladder is 25 ft long

Answer 2

disagree

the equation of the cross-section of the cylinder is
(x-6)^2 + (y-6)^2 = 36
2(x-6) + 2(y-6)^2 dy/dx= 0
dy/dx= (x-6)/(y-6) <--- slope of the tangent = slope of the ladder = -3/4

(x-6) = (-3/4)(y-6)
sub back in the circle equation ...
[(3/4)(y-6)]^2 + (y-6)^2 = 36
(9/16)(y-6)^2 + (y-6)^2 = 36
(y-6)^2 (9/16 +1) = 36
(y-6)^2 = 576/25
y-6 = ± 24/5 , there are two tangents, we want the outer one
y = 24/5 + 6 = 54/5
then subbing back: x = 48/5

equation of tangent:
y = 54/5 = (-3/4)(x - 48/5)
...
15x + 20y = 360
x-ntercept, let y=0 , x = 24
y-intercept, let x = 0, y = 18

ladder^2 = 18^2 + 24^2 = 900
Ladder = 30