A balloon and a bicycle A balloon is rising v...

Question

A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

A hot air balloon rises vertically while a bicycle moves horizontally below it, illustrating related rates in calculus.

Accepted Answer

Hello. In this video, we are going to be approaching the following related rates problem. So we are told that a drone is flying vertically away from a straight and level road at a constant speed of 3 ft per second. At the same moment, a car starts moving on the road directly below the drone at a constant speed of 12 ft per second. If the drone is initially 40 ft above the road, how fast is the distance between the car and the drove increasing 2 seconds later? So, let's go ahead and draw an image to go with this picture. So first, we were told that we have a straight and level road. Now, above the street and level road, we have a drone. The drone's initial height is 40 ft. Now right below the drone, we had a car. Now, this car is traveling at a rate of 12 ft per second. Which means that after some time has passed, the car will be some location away from the drone. Now, the drone is rising at a rate of 3 ft per second. And what we want to do is we want to find the rate of change between the distance and of the car and the drone. So, we need to go ahead and finish filling out this picture. So, what is a shape that we can draw that can relate all the information together? Well, keep in mind that the drone is initially 40 ft above the ground, but we want to find the rate of change of the distance between the car and the drone 2 seconds later. So, its starting position, the drone, is 40 ft, but what is the extra height that it traveled within 2 seconds? Well, since the drone is rising at 3 ft per second, That means the drone would have risen an additional 6 ft across 2 seconds. So, the drone is at a higher position and the additional height that it traveled within 2 seconds is 6 ft because it's traveling at a rate of 3 ft per second, which means that the total height that the drone is at is going to be 46 ft. Now, after traveling 2 seconds, the car's position is going to be some known distance away from the drone, but we can represent the distance between the drone and the car with this diagonal line. What we have here is a right triangle with some hypotenuse, which we will label as C. Now, what is the length of this triangle? Well, since we know that the car is traveling 12 ft per second, we want to find the distance between the drone and the car 2 seconds later, which means that the car would have traveled a total distance of 24 ft within 2 seconds. So the length of this triangle is going to be 24 ft. Now what we need to do is we need to solve for the hippo new C. In order to do this, we will go ahead and use the Pythagorean theorem. C2 is equal to A2 + B2. Now, we know that after 2 seconds, the total height the the drone is at is 46 ft. That is going to be our A. And our B is the length or the distance the car has traveled, which is 24. So we will have C2 is equal to 46 squad. Plus 24 squared. This will simplify to give us 2,692. So we have C squared is equal to 2,692. And once we take the square root of both sides of this equation, we will have C is equal to 2 multiplied by the square root of 673. This is going to be the distance 2 seconds later, between the car and the drone. Now, what we are trying to solve for is we are trying to figure out how fast this distance is changing. So, what we need to do is we need to introduce some type of derivative that is going to give us the rate of change of this characteristic. Well, we can go ahead and start with the Pythagorean theorem again, since this relates to the three sides of the triangle we created. This will give us C2 is equal to A2 + B2. Now what we want to do is we want to introduce some derivatives that describe the rate of change of the sides with respect to time. So we are going to take the derivative with respect to time of this Pythagorean theorem. That is going to give us 2C DC over DT, which is the rate of change of the hypotenuse, equal to 2A multiplied by DA over DT. Which is the rate of change of the height, plus 2 B DB over DT, which is the rate of change of the car. Since we are trying to figure out the rate of change between the car and the drone, we are going to solve this equation for DCDT. We will go ahead and divide 2C to both sides of this equation, leaving us with DCDT equal to 2 A DADT. Plus 2 B D B D T divided by 2 C. Notice that all the terms in this ratio have a coefficient of 2. We can divide the numerator and denominator by 2, and that will allow us to simplify the factors of 2 out of the time derivative. Now what we need to do is we just need to plug in our known information. So A is going to be the total distance traveled by the drone after 2 seconds, that is 46. So we have 46 multiplied by the rate of change of the drone, which is 3, plus The total distance of the car traveled, which is 24, multiplied by the car's rate of change. Which is 12, divided. By the length of the hypotenuse or the distance between the car and the drone, which is 2 square root of 673. The numerator Of this rate of change will simplify to be 426. So we have 426 divided by 2 multiplied by the square root of 673. 426 divides into 2 213 times, so we have 213 divided by the square root of 673. Finally, if we rationalize this this quantity, we will have 213 multiplied by the square root of 673 divided by 673, and this is going to be the number that describes the rate of change of the distance between the drone and the car 2 seconds later. So I hope this video helps you in understanding how to approach this related rates problem, and I will go ahead and see you all in the next video.

Key Concepts

Related Rates

Pythagorean Theorem

Differentiation

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