Calculus question (again)!?

Question

A plank is used to reach over a fence 8 ft high to support a wall that is 1 ft behind the fence. What is the length of the shortest plank that can be used? The plank is the hypotenuse of a triangle formed by the wall and the ground.

How do I even begin approaching this? The hint is that I'm supposed to express the length of the plank in terms of the angle opposite the wall.

Answer 1

Let t = angle opposite the wall
Let P = total length of plank
Let Q = length of plank behind the fence
Therefore, P - Q = length of plank in front of the fence

In the triangle in front of the fence, we have :
sin(t) = 8 / (P - Q)

In the little triangle at the top, behind the fence, we have :
cos(t) = 1 / Q

From the 1st equation, Q = P - 8 / sin(t)
From the 2nd equation, Q = 1 / cos(t)

Equating these 2 values for Q gives :
P - 8 / sin(t) = 1 / cos(t)

From which, we get P in terms of t :
P = 1 / cos(t) + 8 / sin(t)

Differentiating gives :
dP/dt = sin(t) / cos^2(t) - 8cos(t) / sin^2(t)

Equating this to zero gives the shortest P :
sin(t) / cos^2(t) - 8cos(t) / sin^2(t) = 0

Therefore, sin(t) / cos^2(t) = 8cos(t) / sin^2(t)

Cross multiplying gives :
sin^3(t) = 8cos^3(t)

Taking the cube root of both sides gives :
sin(t) = 2cos(t)

Dividing through by cos(t) gives :
tan(t) = 2

Going back a little, we had :
P = 1 / cos(t) + 8 / sin(t)

Multiplying the RHS by sin(t) / sin(t) gives :
P = [1 / sin(t)][tan(t) + 8]

From the formula, csc^2(t) - cot^(t) = 1, we get :
1 / sin(t) = sqrt[tan^2(t) + 1] / tan(t)

Therefore, P = sqrt[tan^2(t) + 1] * [tan(t) + 8] / tan(t)

Substituting 2 for tan(t) gives :
P = sqrt[2^2 + 1] * [2 + 8] / 2
= 5 * sqrt(5) feet, as the shortest plank,
which is approximately 11.2 feet.

Answer 2

Assuming the wall is perpindicular to a level piece of ground, you can figure that angle is 90 degrees. You already have part of the distance of the base that being 1ft. If you label the unknown part x, then the length of the base would be x+12 as expressed in inches. You have the height that you need, 8 ft, or 96 in. Label the hypotenuse y. You have one side at 96, one side at x+12, and one side y, with a right angle of 90 degrees.

I hope this helps. I had calculus and algebra many years ago, but did well in both.

Answer 3

It seems a simple Pysagor problem. Hyp= ((l)^2+(w)^2)^(1/2)

Hyp= (8^2+1^2)^(1/2)=~8.06

You can use sinus and cosinus here too. If you say "a" to the angle with the plank and the ground. tan(a) = 8/1

so a = 82.9 degees

So you can get all your trigonometric equations from there.

And sorry if I couldn't understand question well =))