A ladder 25 feet long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall so that the top of the ladder is sliding down at 3ft/sec, how fast is
the measure of the measure of the acute angle between the ladder and the ground changing when
the bottom of the ladder is 15 feet from the wall?
Since we are looking for the rate of change of the angle, we need
an equation that contains a trig relationship with the angle.
Let the angle at the base of the ladder be θ.
If the height up the wall is y and the distance of the base of the ladder
to the wall is x, then
x^2 + y^2 = 25^2
for the given case, x = 15
15^2 + y^2 = 25^2
y = 20
given: dy/dt = -3 ft/s when x = 15 and y = 20
find dθ/dt at that instant.
I choose
sinθ = y/25 , since we need dθ/dt and we want to involve the given dy/dt
25 sinθ = y
25cosθ dθ/dt = dy/dt
dθ/dt = -3/(25cosθ) = -3/(15/25) = -5
the angle is decreasing at 5 radians/sec
(the rate seems a bit big, better check my arithmetic
Well, well, well, it seems like we've got ourselves a moving ladder situation. Let's break it down, shall we?
We know that the ladder is 25 feet long and that the bottom is being pulled horizontally away from the wall. So we have ourselves a right triangle, with the wall, the ground, and the ladder forming the sides.
Now, we need to figure out how fast the angle between the ladder and the ground is changing. To do that, we need to find the rate of change of that angle.
Let's call the angle we're interested in "θ". The tangent of θ is defined as the opposite side (the height of the wall) divided by the adjacent side (the distance from the wall to the ladder's bottom). So we have:
tan(θ) = height / distance
Now, we want to find dθ/dt, the rate of change of θ with respect to time. To do that, we need to differentiate both sides of the equation with respect to time. But wait, we have a problem. The height isn't changing, so its derivative is 0. What remains is the rate of change of the distance:
sec²(θ) * dθ/dt = d(height) / dt / distance
Simplifying a bit, we get:
dθ/dt = (d(height) / dt) * (1 / (distance * sec²(θ)))
Now, we know that d(height) / dt is given to be -3 ft/sec (since the top of the ladder is sliding down at a rate of 3 ft/sec), and we also know that the distance is 15 feet (as given in the question).
So, let's plug in these values and find our answer:
dθ/dt = (-3 ft/sec) * (1 / (15 ft * sec²(θ)))
Now, we still need to find the value of sec²(θ). To do that, we can use the Pythagorean theorem, since we're dealing with a right triangle. We have:
distance² + height² = ladder length²
15² + height² = 25²
225 + height² = 625
height² = 625 - 225
height² = 400
height = 20 ft
Now, sec²(θ) is defined as (hypotenuse / adjacent)², so we get:
sec²(θ) = (25 ft / 15 ft)²
sec²(θ) = (5/3)²
sec²(θ) = 25/9
Plugging this value back into our previous equation:
dθ/dt = (-3 ft/sec) * (9/25)
dθ/dt = -27/25 ft/sec
And there you have it! The measure of the acute angle between the ladder and the ground is changing at a rate of -27/25 ft/sec.
To solve this problem, we can use trigonometry and related rates.
Let's denote the distance between the bottom of the ladder and the wall as x, and the length of the ladder as L.
We know that the top of the ladder is sliding down at a rate of 3 ft/sec. Let's denote the rate at which the distance is changing as dy/dt.
To find dy/dt, we need to find the relationship between x and y (the height of the ladder on the wall).
Using the Pythagorean theorem, we have:
x^2 + y^2 = L^2
Differentiating both sides with respect to time t gives:
2x(dx/dt) + 2y(dy/dt) = 0
Rearranging this equation gives:
dy/dt = -x(dx/dt)/y
We are given dx/dt = 3 ft/sec and x = 15 ft (when the bottom of the ladder is 15 ft from the wall). We need to find dy/dt when y is unknown.
Using the Pythagorean theorem again with x = 15 ft, we have:
15^2 + y^2 = 25^2
225 + y^2 = 625
y^2 = 400
y = 20 ft
Now, substituting the known values into the equation for dy/dt:
dy/dt = -15(3)/(20)
dy/dt = -45/20
dy/dt = -9/4 ft/sec
Therefore, the measure of the acute angle between the ladder and the ground is changing at a rate of -9/4 ft/sec when the bottom of the ladder is 15 feet from the wall.
To find the rate at which the acute angle between the ladder and the ground is changing, we can use trigonometry and related rates.
Let's denote the distance between the bottom of the ladder and the wall as x, and the height of the ladder from the ground as y. The ladder forms a right triangle with the wall and the ground.
Using the Pythagorean theorem, we have:
x^2 + y^2 = 25^2
Now, let's differentiate both sides of the equation with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
Since we're interested in finding how fast the acute angle is changing, we can focus on dy/dt.
Given that dx/dt = 3 ft/sec (the rate at which the bottom of the ladder is pulled horizontally away from the wall) and we want to find dy/dt (the rate at which y is changing), we can substitute these values into the equation:
2x(3) + 2y(dy/dt) = 0
Now, we need to find the values of x and y when the bottom of the ladder is 15 feet away from the wall.
In the right triangle, we can use x as the adjacent side and y as the opposite side of the acute angle we're interested in. Using trigonometry, we have:
tan(theta) = y/x
=> theta = arctan(y/x)
When x = 15 ft, we can substitute this value into the equation above to solve for y.
x^2 + y^2 = 25^2
15^2 + y^2 = 25^2
y^2 = 625 - 225
y^2 = 400
y = 20 ft
Now, we can substitute these values into the equation we derived earlier:
2(15)(3) + 2(20)(dy/dt) = 0
Simplifying the equation:
90 + 40(dy/dt) = 0
40(dy/dt) = -90
dy/dt = -90/40
dy/dt = -9/4 ft/sec
Therefore, the measure of the acute angle between the ladder and the ground is changing at a rate of -9/4 ft/sec when the bottom of the ladder is 15 feet from the wall.