Constant Rate of Change05:21 minutes

Video Transcript

TranscriptConstant Rate of Change

Giovanni, an Italian poet and a very old-school kind of guy, wants to send a letter to his sweetheart. The only problem is his swain lives in the neighboring kingdom, 250 miles away. He visits the local carrier pigeon salesman and purchases the fastest pigeon in all the land. Giovanni's quite impatient and wants to know when the pigeon will reach his paramour. To figure this out, and play Medieval matchmaker for Giovanni, we can use constant rate of change. When something has a constant rate of change, one quantity changes in relation to the other. For example, for every half hour the pigeon flies, he can cover a distance of 25 miles. We can write this constant rate as a ratio. For ratios, it's always a good idea to state both units in whole terms. Since we don't want to change the ratio, we multiply the ratio by its multiplicative identity property. In this case, that means multiplying both the numerator and denominator by 2 to get a whole number in the denominator. And, just like with fractions, you should always write ratios in lowest terms. Simplified, the constant rate is 50 miles per hour. Let's help Giovanni figure out where his pigeon will be after a certain amount of time has passed. We can create a table using different values for time to see how much progress the pigeon has made. We know the pigeon flies at a rate of 50 miles per hour, so after one hour, the pigeon will have flown a total of 50 miles. To find out how far the pigeon will have flown after 2 hours, we just have to multiply the rate by 2. If the pigeon flies for 3 hours, we can simply multiply our original rate by 3 to find out that the pigeon will have traveled 150 miles. We can continue this process, filling in our table as we go. Okay. Looking good. But, to get an even better understanding, let's graph the data from the table. Let the x-axis represent the elapsed time in hours, and let the y-axis represent the distance in miles. Each 'x-y', pair represents a location on the graph. Just like 'x' is an independent variable, the number of hours is the independent variable in this example. So, the total distance, 'y', is the dependent variable. Now plot the points from the table onto the graph using the ordered pair (hours and miles). Notice how the graph is a perfectly straight line! Notice that, for every hour, the pigeon flies an additional 50 miles. We can also solve this problem algebraically. Let's go back to the equation we set up earlier. We found out earlier that the rate the pigeon flies is 50 miles per hour. Substituting this back into our equation, we have rate times 2 hours equals 100 miles. Since an hour is a unit of time, we can substitute the '2 hours' in the equation with 'time'. And finally, 100 miles is the distance the pigeon flew, so we can substitute '100 miles' with 'distance'. Mathematicians typically write equations with single variables instead of words. So, to make it shorter to write, let's just use the first letters of each word. Since distance, or 'd', is our dependent variable, it's customary to write it on the left side of the equation. There, that's better. Now, we can solve any constant rate problem if we know just two of these variables. Let's plug in what we know and find out where that plumy pigeon is. He needs to fly a distance of 250 miles at a constant rate of 50 miles per hour. Now, multiplying by the reciprocal of our rate to isolate the 't' and solve Giovanni's little problem, it'll take the pigeon 5 hours to deliver the letter. Giovanni waits and waits for a response from his sweetheart. I guess he’ll just have to wait a little while longer.

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