A 25ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle between the ground and the ladder changing, if the top of the ladder is sliding down the wall at the rate of r inches per second, at the moment that the top of the ladder is h feet from the ground. (equation in terms of h and the angle)
a. r / (300√ (1 - (h/25)^2))
b. 1 / (25√ (1 - (h/25)^2))
c. r / (12√ (1 - (h/25)^2))
d. r / (25√ (1 + (h/25)^2))
e. 1 / (25√ (1 + (h/25)^2))
It's A 10/10 cannot recommend B
am I correct
Well, well, well. Let's see if I can solve this equation with a touch of humor!
First, let's define our variables. We have r, the rate at which the top of the ladder is sliding down the wall, and h, the height of the ladder from the ground.
We need to find the rate at which the angle between the ground and the ladder is changing. To do this, we can use some trigonometry.
Let's call the angle θ (because I like Greek letters). We know that the tangent of θ (tan(θ)) is equal to h/25, right? Dividing the opposite side (h) by the adjacent side (25ft).
Now, let's differentiate both sides of this equation with respect to time (t). Don't worry, I'm not going to differentiate your personality or anything!
We get sec²(θ) * dθ/dt = (1/25) * dh/dt.
To solve for dθ/dt, we can rearrange the equation:
dθ/dt = (1/25) * dh/dt * (1/sec²(θ)).
But we need the equation in terms of h and the angle, not θ. Luckily, we know that sec²(θ) is equal to 1 + (h/25)², thanks to the Pythagorean identity.
So, we can substitute sec²(θ) with (1 + (h/25)²):
dθ/dt = (1/25) * dh/dt * (1/(1 + (h/25)²)).
Simplifying the equation a bit:
dθ/dt = (1/25) * dh/dt / √(1 + (h/25)²).
And that's our answer! Option (e) is the closest to the correct equation: 1 / (25√ (1 + (h/25)²)).
Remember folks, math can be fun! Just add a dash of humor and you'll have everyone laughing with you.
To solve this problem, we can use trigonometry and related rates. Let's first set up some variables:
Let θ be the angle between the ground and the ladder.
Let h be the height of the top of the ladder above the ground.
Let r be the rate at which the top of the ladder is sliding down the wall.
We're asked to find an equation that relates θ, h, and r.
Since we have a right triangle formed by the ladder, the wall, and the ground, we can use trigonometry. In particular, we can use the sine function:
sin(θ) = h/25
To relate θ, h, and r, we need to find an equation that involves these three variables. To do this, we'll differentiate the above equation with respect to time. Keep in mind that h and r are both changing with respect to time, so we need to use the chain rule.
Differentiating both sides of the equation gives us:
cos(θ) * dθ/dt = (1/25) * dh/dt
Now, we rearrange this equation to solve for dθ/dt, the rate at which the angle is changing:
dθ/dt = (1/25) * (dh/dt) / cos(θ)
Since we are given that the top of the ladder is sliding down the wall at a rate of r inches per second (dh/dt = -r), we substitute this:
dθ/dt = (1/25) * (-r) / cos(θ)
Next, we substitute the value of cos(θ) from our original equation sin(θ) = h/25:
dθ/dt = (r/25) / cos(θ) = r / (25 * cos(θ))
To relate this to the answer choices, we need to express the equation in terms of h instead of θ. We can do this by substituting cos(θ) with √(1 - sin^2(θ)).
Recall that sin(θ) = h/25:
cos(θ) = √(1 - (h/25)^2)
Now, substitute this back into the equation:
dθ/dt = r / (25 * √(1 - (h/25)^2))
Comparing this to the answer choices, we see that the correct option is:
a. r / (300 * √(1 - (h/25)^2))
Therefore, the rate at which the angle between the ground and the ladder is changing is r / (300 * √(1 - (h/25)^2)).
sinθ = h/25
cosθ dθ/dt = 1/25 dh/dt
dθ/dt = -r/25 * 25/√(625-h^2) = -r/√(625-h^2) = -r/(25√(1 - (h/25)^2)
Looks like B, except that the numerator should be -r, not 1, since clearly the angle is decreasing as the ladder slides down.