Question 1.
A stuntman falls from a hot air balloon that is hovering at a constant altitude,100 feet above a lake. A television camera on shore, 200 feet from a point directly below the balloon,follows his descent. At what rate is the angle of elevation, theta, of the camera changing 2 seconds after the stuntman falls? ( neglect the height of the camera)
It seems like I'm missing a rate of change to solve this problem. I think I need to do something with the 2 seconds,but I don't know what. Any help would really be appreciated.
Question 2
An isosceles triangle has equal sides of length 12m. If the angle theta between these sides is increased from 30 to 33 degrees, use differentials to approximate the change in the area of the triangle.
I know how to find the area, missing sides, and the height. My intial thought was to do a derivative of the area and multiply by 3 degrees. However, i don't have any given rates so I am lost on what to do. Any help would really be appreaciated.
The length of half the base is 12 sin(θ/2), so the area of the triangle is
a = 1/2 bh = 1/2 (24 sin(θ/2)) (12 cos(θ/2))
= 144 sinθ
da = 144 cosθ dθ
θ increases by 3θ, so dθ = π/60
da ~= 144 cos π/6 * π/60 = 144 * √3/2 * π/60 = 6π/5 √3
Question 1:
To solve this problem, you need to use trigonometric functions and relate the given rates of change.
Let's start by drawing a diagram. We have a right triangle where the balloon is at the top vertex, the point directly below the balloon is at the bottom vertex, and the camera is at the third vertex. The distance from the camera to the balloon is 200 feet, and the height of the balloon is 100 feet. We are interested in finding the rate at which the angle of elevation, theta, is changing 2 seconds after the stuntman falls.
To find this rate, we will use the relationship between the tangent function and the angle of elevation. The tangent of an angle is equal to the opposite side divided by the adjacent side. In this case, the tangent of theta is the height of the camera (200 feet) divided by the distance between the camera and the point directly below the balloon (x feet). So we have:
tan(theta) = 200 / x
To find the rate at which theta is changing, we need to differentiate this equation with respect to time (t) since we are given a time. Differentiating both sides of the equation gives us:
sec^2(theta) * d(theta)/dt = -200 / x^2 * dx/dt
Here, d(theta)/dt represents the rate of change of theta with respect to time.
We are given that x is the horizontal distance between the camera and the point directly below the balloon. Since the stuntman is falling vertically, we can assume that the horizontal distance x is changing at a constant rate. We are given that dx/dt = 0 since x is not changing. Therefore, the right side of the equation becomes 0.
So we are left with:
sec^2(theta) * d(theta)/dt = 0
Now, solve this equation for d(theta)/dt to find the rate at which the angle of elevation is changing. Since the secant squared function is never zero, the rate d(theta)/dt must be zero. Therefore, the angle of elevation is not changing 2 seconds after the stuntman falls.
Question 2:
To approximate the change in the area of the triangle when the angle theta changes from 30 to 33 degrees, we can use calculus and differentials.
The formula for the area of an isosceles triangle is given by A = (1/2) * base * height. In this case, the base of the triangle is 12m and the height can be found using trigonometry.
Let's set up the equation for the area of the triangle:
A = (1/2) * base * height
Now, let's find the derivative of both sides of the equation:
dA = (1/2) * (dbase * height + base * dheight)
Since we are interested in the change of the area, we can substitute dA with ΔA to represent the change in the area. Similarly, we can use Δbase to represent the change in the base and Δheight to represent the change in the height. Therefore, the equation becomes:
ΔA = (1/2) * (Δbase * height + base * Δheight)
We are given that the equal sides of the triangle have a length of 12m, so the base is constant. Therefore, Δbase is equal to zero.
Now, we need to find the value of Δheight. We are given that the angle theta between the equal sides changes from 30 to 33 degrees.
To find the relationship between the base, height, and angle theta, we can use trigonometry. In this case, the height of the triangle is the side opposite to the angle theta, and the base is the adjacent side. Since we know the length of the equal sides is 12m, we can use the sine function to relate the height and the angle theta:
sin(theta) = height / base
We can rearrange this equation to solve for the height:
height = base * sin(theta)
Now, let's substitute this into our equation for ΔA:
ΔA = (1/2) * (0 + 12m * Δheight) = 6m * Δheight
To find the value of Δheight, we can use differentials. We know that the angle theta changes from 30 to 33 degrees, so Δtheta is equal to 3 degrees.
Differentiating the equation height = base * sin(theta) gives us:
dheight = base * cosine(theta) * dtheta
Substituting the given values, we have:
dheight = 12m * cos(30 degrees) * 3 degrees
Now we can approximate Δheight using the differential:
Δheight = dheight = 12m * cos(30 degrees) * 3 degrees
Finally, substitute this value into our equation for ΔA:
ΔA = 6m * Δheight = 6m * (12m * cos(30 degrees) * 3 degrees)
Now you can calculate the approximate change in the area of the triangle when the angle theta increases from 30 to 33 degrees using the given values.
the height of the stuntman at time t seconds after falling is
y = 100-16t^2
dy/dt = -32t
When the stuntman is at height y,
tanθ = y/200
sec^2 θ dθ/dt = 1/200 dy/dt
(1+tan^2θ) dθ/dt = 1/200 dy/dt
(1+(y/200)^2) dθ/dt = 1/200 dy/dt
at t=2, y=36, dy/dt = -64, so
(1+(36/200)^2) dθ/dt = 1/200 (-64)
dθ/dt = -.3 rad/s or -17.8°/s