A 15ft ladder leans against a building and its base starts to slide down the wall. The base is moving at a rate of 5ft/sec when the base is 9ft from the wall. How fast is the top of the ladder moving at this moment? What is the rate of change of the angle formed by the ladder and the ground?
I know that this forms a right triangle with a hypotenuse of 15 an two legs that measure 9 (for x) and 12 (for y).
My first step was:
x dx/dt + y dy/dt = z dz/dt
9(5) + 12 dy/dt = 15(0)
dy/dt = 3.75 ft/sec
Then I attempted to answer the second part of the question, but I wasn't sure how to do this.
What I did was:
A=1/2bh or 1/2xy
dA/dt = 1/2 x dy/dt + 1/2 y dx/dt
dA/dt = 1/2(9)(3.75) + 1/2(-12)(5)
dA/dt = -105/8 ft^2/sec
-------> I'm not sure if this is the right process or if I'm supposed to be looking for an angle? Any help is appreciated, thanks
Your first part is correct, except your
dy/dt should be -3.75 (showing that is slides "down")
for your second part , since we are involving an angle, we have to introduce an equation containing an angle.
Trig will do this.
So let the base angle be Ø
cosØ = x/15
15cosØ = x
-15sinØ dØ/dt = dx/dt
for our given case, sinØ = 12/15 = 4/5
dØ/dt = -5/(15(4/5)) = -5/12 radians per
The negative says that the angle is decreasing, which makes sense as the ladder is falling.
To find the rate at which the top of the ladder is moving, we can use a similar approach to what you did for the first part of the question. Let's assign the variables as follows:
- x represents the distance of the base from the wall
- y represents the height of the ladder
- z represents the length of the ladder (constant)
Given that dz/dt = 0 (the length of the ladder is not changing), we can use the Pythagorean theorem to relate x and y:
x^2 + y^2 = z^2
Differentiating both sides of the equation with respect to time, we have:
2x*dx/dt + 2y*dy/dt = 2z*dz/dt
Since dz/dt is 0, the equation simplifies to:
x*dx/dt + y*dy/dt = 0
Substituting the known values, x = 9 ft and dx/dt = 5 ft/sec, we can solve for dy/dt:
9 * 5 + 12 * dy/dt = 0
45 + 12 * dy/dt = 0
12 * dy/dt = -45
dy/dt = -45/12
dy/dt = -3.75 ft/sec
Therefore, the top of the ladder is moving down at a rate of 3.75 ft/sec.
Now, for the rate of change of the angle formed by the ladder and the ground, let's call it θ. We can use trigonometry to relate x, y, and θ:
tan(θ) = y/x
Taking the derivative of both sides with respect to time, we get:
sec^2(θ) * dθ/dt = (x * dy/dt - y * dx/dt) / (x^2)
Substituting the known values, x = 9 ft, y = 12 ft, dx/dt = 5 ft/sec, and dy/dt = -3.75 ft/sec:
sec^2(θ) * dθ/dt = (9 * (-3.75) - 12 * 5) / (9^2)
sec^2(θ) * dθ/dt = (-33.75 - 60) / 81
sec^2(θ) * dθ/dt = -93.75 / 81
dθ/dt = -93.75 / (81 * sec^2(θ))
So, the rate of change of the angle is given by dθ/dt = -93.75 / (81 * sec^2(θ)).
Note that sec^2(θ) is the square of the secant function, which relates to the angles. You need to know the value of θ to compute the rate of change of the angle accurately.