A ladder is intended to reach the wall of a building, but it must clear a fence with the length of d that stands at the distance b from
that wall. Find the shortest length of ladder required to arrange this.
note use differential calculus
Draw a diagram. Let
x = length of ladder
z = distance from fence to base of ladder
z/√(b^2+z^2) = (z+d)/x
xz = (z+d)√(b^2+z^2)
x = (z+db^2)√(b^2+z^2)/z
dx/dz = (z^3-db^2)/(z^2 √(b^2+z^2))
dx/dz=0 when z^3 = db^2
Use that to evaluate x
To find the shortest length of ladder required to clear the fence and reach the wall, we can use differential calculus to find the minimum distance.
Let's assume that the ladder makes a right angle triangle with the ground (base), the wall (height), and the ladder itself (hypotenuse). The fence forms the other side of the triangle.
Let's label the hypotenuse as L, the height as h, and the base as x.
Using the Pythagorean theorem, we know that:
L^2 = b^2 + x^2 ...(1)
We also know that the ladder must reach the wall, so:
h = L
We need to minimize L. To do this, we can differentiate equation (1) with respect to x and find the critical point where the derivative equals zero.
Differentiating equation (1) with respect to x, we get:
2L * dL/dx = 2x
Now, solving for dL/dx:
dL/dx = x/L
Since we know that h = L, we can rewrite this as:
dL/dx = x/h
Setting dL/dx equal to zero to find the critical point, we have:
x/h = 0
This implies that x = 0. Thus, at x = 0, we have a critical point.
Now, let's check the second derivative of equation (1) to confirm whether this critical point is a minimum.
Differentiating the derivative dL/dx with respect to x, we get:
d^2L/dx^2 = d(x/h)/dx = 1/h
At x = 0, the height h is positive, so 1/h is also positive. This confirms that we have a minimum at this critical point.
Now, we can find the minimum L by substituting x = 0 back into equation (1):
L^2 = b^2 + 0^2
L = sqrt(b^2)
L = b
Therefore, the shortest length of ladder required to clear the fence and reach the wall is equal to the length of the fence, which is b units.