Related Rates. Problem Example 2
Car Ais 40 miles west of of an intersection driving toward)it at 60 mph. Car B is 30 miles south of the same intersection driving away from it at 45 mph. At what rate is the distance between the two cars changing at this time? [Graph] d/dt[A^2+B^2=C^2] | A=40 2A dA/dt+2B dB/dt=2c dC/dt | B=30 40*(-60)+30*45=50 dC/dt | C=50 -2400+1350=50 dC/dt | dA/dt=-60 dC/dt=-21 mph | dB/dt=45 | dA/dt=? Okay, we’ll call this one an intersection question. We have two cars, A and B. Car A is 40 miles west of an intersection nd traveling toward it at a certain velocity. Car B is 30 miles south of the same intersection and driving away from it at a certain velocity. We want to know how quickly the distance between them is changing. First, I draw an intersection to visualize the situation. Car A, which we will color red, is traveling toward the intersection at 60 miles per hour. He is currently 40 miles away from the intersection. And then car B. At this moment, car B is 30 miles away from the intersection. His velocity is 45 miles per hour. So I want to know how fast the distance between them is changing at this time. You will have to develop your own equation to solve this problem. Here we have created a right triangle, so Pythagorean theorem might be a good choice since it is a formula which can be used to find the length of the third side of right triangles. So if A squared plus B squared equals C squared, then the derivative of that equation will help you figure out what variables are in play. The derivative of A squared is 2A dA dt, where A is a function of x. The derivative of B squared is 2B db dt, where B is a function of x. The derivative of C squared is 2C dC dt, where C is a function of x. It is clear from this list that we need a great deal of supplies. So we have an A, a B, a C, a dA/dt, a dB/dt and a dC dt. There are several ways to solve this problem. We can use the Pythagorean theorem to find C, or we can realize that B is 30 miles away and A is 40 miles away, which forms a right triangle with legs of 3 and 4, so we know that C is 5 miles away from the intersection. Now let us consider the rates of change. If a car is 40 miles west of an intersection and is driving toward it at 60 miles per hour, then its distance from the intersection will decrease. If I am driving toward the intersection, then the distance between the intersection and me is decreasing. That means that my rate of approach, dA/dt, is negative 60. If B is moving away from the intersection, then its distance from the intersection is increasing. This means that db/dt, the rate of change of B, is positive 45. The difference between the rate of change of these two IS equal to our unknown variable and can be found by taking the derivative of A with respect to time, dividing through by 2 and evaluating. Now we have that A is 40. dA/dt is negative 60 plus 30 times 45, which equals 50 dC/dt. And you find that the correct answer is negative 21 miles IS per hour. The distance is decreasing, but if they ask us about it, we’ll have to say that the rate of change is negative 21 miles per hour. Because the derivative of this equation is negative, the distance between the two curves 15 decreasing. The distance between the two objects is decreasing, which indicates that their rates of change with respect to time are negative that is, one object is moving away from the other. Word amount: 579
