# Math Help - integrals of trigonometric functions

1. ## integrals of trigonometric functions

a thief tries to enter a building by placing a ladder over a 9-ft-high fence so it rests against the building,which is 2 ft back from the fence.

what is the length f the shortest ladder and the ground? (let Q be the angle between the ladder and the ground) express the length of the ladder in terms of Q,

and then find the value of Q that minimizes the length of the ladder.

2. let $x$ = distance from foot of the ladder to the fence

$f\cos{Q} = x+2$

$\cot{Q} = \frac{x}{9}$

$f\cos{Q} = 9\cot{Q} + 2$

$f = 9\csc{Q} + 2\sec{Q}$

find $\frac{df}{dQ}$ and minimize $f$

3. after I find out the derivetive, how can I minimize the f? let that equals to 0,but I don't know how to solve the value for Q

4. $
f = 9\csc{Q} + 2\sec{Q}
$

$
\frac{df}{dQ} = -9\csc{Q}\cot{Q} + 2\sec{Q}\tan{Q} = 0
$

$\frac{-9\cos{Q}}{\sin^2{Q}} + \frac{2\sin{Q}}{\cos^2{Q}} = 0$

$
\frac{-9\cos^3{Q}}{\sin^2{Q}\cos^2{Q}} + \frac{2\sin^3{Q}}{\sin^2{Q}\cos^2{Q}} = 0
$

$
-9\cos^3{Q} + 2\sin^3{Q} = 0
$

$
9\cos^3{Q} = 2\sin^3{Q}
$

$
\frac{9}{2} = \frac{sin^3{Q}}{\cos^3{Q}}
$

$\frac{9}{2} = \tan^3{Q}$

$\sqrt[3]{\frac{9}{2}} = \tan{Q}$

$Q = \arctan\left(\sqrt[3]{\frac{9}{2}}\right)$

$
Q \approx 58.8^{\circ}
$

now go back and determine the length of the ladder.