A 25 ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle theta 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? (You're looking here for an equation in terms of h and theta.)
You can find a visual on Chegg by searching the question.
Please help ASAP.
Your sketch should consist of a right-angled triangle.
Hypotenuse = 25 and height = h
angle ladder makes with ground = θ
The given units refer to hypotenuse and opposite side, so
we'll go for sinθ
sinθ = h/25
25sinθ = h
25 cosθ dθ/dt = dh/dt , but we are given that dh/dt = -r inches/s = -r/12 ft/s
dθ/dt = (-r/12)/(25cosθ) = -r/(300cosθ) radians/s
Please look at the first visual from the Chegg website, by searching this question:
Which option do u think is correct?
To find an equation in terms of h and theta, we can start by considering a right triangle formed by the ladder, the ground, and the wall.
Let theta be the angle between the ground and the ladder. Let h be the distance of the top of the ladder from the ground. And let r be the rate at which the top of the ladder is sliding down the wall.
According to the problem, we know that the ladder has a length of 25 ft. Therefore, the hypotenuse of the right triangle is 25 ft.
Using the Pythagorean theorem, we can write the equation:
h^2 + r^2 = 25^2
Differentiating both sides of the equation with respect to time t, we get:
2h * dh/dt + 2r * dr/dt = 0
Since we're interested in finding the rate of change of theta, we need to relate theta to h and r. From the right triangle, we know that:
sin(theta) = h / 25
Differentiating both sides of the equation again with respect to time t, we get:
cos(theta) * d(theta)/dt = (1/25) * dh/dt
Rearranging the equation, we have:
d(theta)/dt = (1/25) * (dh/dt) / cos(theta)
Therefore, the rate of change of theta with respect to time is given by:
d(theta)/dt = (1/25) * (dh/dt) / cos(theta)
Finally, substituting the value of dh/dt from the first equation above, we get:
d(theta)/dt = (1/25) * (-2h * dr/dt) / cos(theta)
So, the equation in terms of h and theta for the rate of change of theta with respect to time is:
d(theta)/dt = (-2h/25) * (dr/dt) / cos(theta)
To solve this problem, we need to make use of trigonometry and related rates.
Let's assume that the angle between the ground and the ladder is theta (θ). We are given that the top of the ladder is sliding down the wall at the rate of r inches per second.
Based on the given information, we can form a right triangle with the wall, the ground, and the ladder as its sides. The ladder serves as the hypotenuse of the triangle.
Here's the equation we can use to relate the changing angle theta (dθ/dt) with the sliding rate of the top of the ladder (dr/dt):
tan(θ) = opposite/adjacent = h/25
Now, we need to differentiate both sides of the equation with respect to time (t):
d/dt(tan(θ)) = d/dt(h/25)
Using the chain rule, we can find the derivatives:
sec^2(θ) * dθ/dt = (1/25) * dh/dt
Now, we solve for dθ/dt:
dθ/dt = (dh/dt * 25) / sec^2(θ)
Remember that sec(θ) is equal to 1/cos(θ). Replacing sec(θ) with its equivalent, the equation becomes:
dθ/dt = (dh/dt * 25) / (1/cos^2(θ))
Simplifying further:
dθ/dt = (dh/dt * 25 * cos^2(θ))
And that is the equation in terms of h and theta that relates the rate of change of the angle theta with respect to time to the sliding rate of the ladder's top.
Please note that the equation may look different in different forms depending on how you simplify it further or substitute values.