A hot air balloon rising straight up form a level field is tracked by a range finder 500ft from the liftoff point. At the moment the range finders elevation ange is pi/4, the angle is increasing at 0.14 rad/min. How fast is the balloon rising at that moment?
h/500 = tan Ø
h = 500 tan Ø
dh/dt = 500sec^2 Ø dØ/dt
sub in the given information
the answer is 140 m/sec
you can view the answer at
this is a website, it wont let me post
hutch.k12.mn.us/userFiles/
File/4.6_Related_Rates_Day_1_c
.pdf
To solve this problem, we can use trigonometry and differentiation.
Let's consider the situation. The range finder is located 500 ft from the liftoff point of the balloon. The elevation angle, denoted by θ, is changing with time. We are given that θ = π/4 (45 degrees), and the angular velocity of the elevation angle is dθ/dt = 0.14 rad/min.
To find the rate at which the balloon is rising, we need to find the derivative of the height of the balloon with respect to time.
Let h(t) be the height of the balloon at time t. We can apply trigonometry to relate the height to the elevation angle:
tan(θ) = h(t) / 500
Differentiating both sides of the equation with respect to time t, we get:
sec^2(θ) * dθ/dt = dh(t)/dt / 500
Since we are given that θ = π/4, we can substitute these values and simplify the equation:
sec^2(π/4) * 0.14 = dh(t)/dt / 500
1.4142^2 * 0.14 = dh(t)/dt / 500
2 * 0.14 = dh(t)/dt / 500
0.28 = dh(t)/dt / 500
Solving for dh(t)/dt, we get:
dh(t)/dt = 0.28 * 500
dh(t)/dt = 140 ft/min
Therefore, the balloon is rising at a rate of 140 ft/min at the given moment.
To find how fast the balloon is rising at that moment, we need to use the concept of related rates. Related rates involve the derivatives of two or more variables that are related to each other. In this case, the variables are the distance from the liftoff point (r) and the elevation angle (θ).
Let's start by noting the given information:
- Range finder distance (r) = 500 ft
- Elevation angle (θ) = π/4
- Rate of change of elevation angle (dθ/dt) = 0.14 rad/min
We need to find the rate of change of the distance (dr/dt) with respect to time when θ = π/4.
To relate the variables r and θ, we can use trigonometry. Recall that the tangent function (tan) is defined as the ratio of the opposite side to the adjacent side of a right triangle.
In this case, the opposite side is the height (h) of the balloon, and the adjacent side is the horizontal distance (x) from the balloon to the range finder. Therefore, we have:
tan(θ) = h / x
Now, we can differentiate both sides of this equation with respect to time (t), keeping in mind that h is the variable we want to solve for:
sec^2(θ) * dθ/dt = dh/dt / x
We can rearrange this equation to solve for dh/dt:
dh/dt = sec^2(θ) * dθ/dt * x
Now, we need to find x when θ = π/4. From trigonometry, we know that x = r tan(θ), so when θ = π/4:
x = r tan(π/4) = r
Therefore, we can substitute x = r into the equation:
dh/dt = sec^2(θ) * dθ/dt * r
Plugging in the given values:
- sec^2(π/4) = 2
- dθ/dt = 0.14 rad/min
- r = 500 ft
dh/dt = 2 * 0.14 * 500 ft/min
Simplifying the equation:
dh/dt = 140 ft/min
Therefore, the balloon is rising at a rate of 140 ft/min at that moment.