Math 19: Calculus Winter 2008 Instructor: Jennifer KlokeLecture Outline (Related Rates)Wednesday, March 5Announcements1. Final is on Monday, 3/17/08 at 8:30 AM. If you can’t make it then, the only other timeyou can take it is at the University scheduled make-up time which is M onday nightfrom 7-10 (but you have a very good excuse to take it at this time.)2. The last day of class (3/14/08) will be a review session for the final - come preparedwith questions.Motivation for Related RatesAs we saw last time, there are many applications of implicit differentiation. Today andtomorrow we are going to use it to talk about very concrete applications called related rates.To illustrate the idea of related rates, consider a balloon (we’ll assume it is shaped likea sphere) being filled with air. Let’s consider two quantities: the volume of the balloonand the radius of the balloon. Of course, we know how they are related: V = 4/3πr3(the volume of a sphere). Except that in this problem, we have another, hidden variable:the time t. Because as time progresses, both the radius and the volume are functions of time.If we are using a machine to pump air into the balloon, there is a good chance we knowthe rate at which the air is being pumped in. The rate at which the air is being pumped inis precisely the rate of change of the volume (with respect to time) which isdVdt. But whatabout the radius? At what rate is the radius changing? That is, what isdrdt? The trick is tolook at the equation V =43πr3and implicitly differentiate with respect to t! We getdVdt= 4πr2drdt.Suppose the problem said that the air was being pumped in at 10 cm3/sec and asked forthe rate at which the radius was changing when the radius was 10 cm. Then1 = = 4π(10)2drdtdrdt=140πSo the radius is changing at140πcm/sec when the radius is 10 cm.1Theory: Related RatesThe strategy for solving related rates is:1. Read the problem carefully.2. Introduce notation. Assign variables to all quantities that are functions of time.3. Draw a picture if possible.4. Express the given information and the required rate in terms of derivatives.5. Write an equation that relates teh various quantities of the problem. If necessary, usethe geometry of the situation to eliminate one of the variables by substitution (seeExample 3 in Section 4.1.)6. Use the chain rule to (implicitly) differentiate both sides of the quation with respect tot.7. Substitute the given information into the resulting equation and solve for the unknowrate.Example: A Related Rate ProblemProblem: A plane flying horizontally at an altitude of 1 mile and a speed of 500 miles/hourpasses directly over a radar station. Find the rate at which the distance from the plane tothe station is increasing when it is 2 miles away from the station.Discussion: After reading the problem carefully, the next few steps in a related ratesproblem are to draw a diagram, introduce variables, write down what we know in terms ofthe variables, and se t up an equation relating the variables. As we introduce variables, we’llgo back to the diagram and fill it out more completely.The key point here is that there are three different types of quantities in any related ratesproblem. Namely:1. Quantities that do not change at all (ones that are the same no matter what the time.)2. Quantities that vary depending on time.3. Quantities that are only true at a specific moment in time.Solution: Consider the above problem. First we are told that the plane has an altitudeof 1 mile. That is true no matter what the time, so that is a quantity of type 1. Secondly,we are told the plane has a speed of 500 miles per hour; this is also true no matter whatthe time, and is a quantity of type 1. The problem mentions the distance from the plane tothe station. This changes depending on the time, so it is a quantity of type 2. Finally theproblem mentions the plane being 2 miles away. This is only true at a specific moment in2time, and hence is of type 3.So what do we do with different types of quantities? First you should assign a variable toevery quantity of type 2; after all, each such quantity varies depending on time, and shouldbe a variable. In our problem, the distance to the station is such a quantity: let’s call it d.Secondly, we can label the diagram with the quantities of type 1 and 2 (we want ourpicture to be true no matter what time it is, so we don’t use any quantities of type 3). Inour problem, we label altitude of the plane as 1 mile, the distance from teh plane to thestation as d, and I would draw a little arrow in the direction the plane is going and label it500 miles/hour.Also, if some quantity of type 1 is a rate of change, it should be the derivative of a vari-able (a quantity of type 2). Now the speed of the plane is how fast the horizontal distancebetween the station and the plane is changing. We don’t have a variable for that, but wenow see that we s hould. Let’s call it x and label it in our diagram (it’s a quantity of type2). Then we know thatdxdt= 500.Finally, before setting up any equations, let’s notice that what we are trying to find isdddtwhen the plane is 2 miles away.Looking back at the diagram, we realize that we have a right triangle, and can write:x2+ 1 = d2.This equation is true, but it’s not immediately helpful since what we really need tofind isdddt, and so we need an equation that involvesdddt. We can get such an equation bydifferentiating the equation above with respect to t.2x(dxdt) = 2d(dddt)dddt=xd(dxdt)Whoop! Now we know thatdddtis always equal toxd(dxdt). We know thatdxdtis always 500,so let’s substitute that in to find thatdddtis always equal to 500x/d.If we just wanted to know whatdddtwas in general, this would be the answer. B ut we wantto know whatdddtis at a specific moment in time, namely when the plane is 2 miles away fromthe station. This is when our quantities of type 3 come into play. At that specific momentin time, we know that d = 2. Sodddt= 250x. But we want a numerical answer, so we’d betterfigutre out what x is when the plane is 2 miles away from the station. Fortunately, we knowthat d2= x2+ 1, so if d = 2, we can solve to find that x =√3. Plugging that in, we finallyfind thatdddt= 250√3 when the plane is 2 miles away from the station. That is, when theplane is 2 miles away from the station, the distance between the plane and the station ischanging (increasing) at 250√3 which is about 433 miles per hour.3More examples1. Gravel is dumped from a