Math 142 (Summer 2018) Business Calculus 5.8 Notes Implicit Differentiation and Related Rates Why? We have learned how to take derivatives of functions, and we have seen many applications of this. However sometimes a function is not described by an explicit rule. These are called implicit functions. In this section we learn how to take derivatives of these functions. We see how to apply this to solve problems involving related rates. Implicit functions and their derivatives Consider the equation x 2 + y 2 = 1. This equation can be used to define y as a function of x. If we set x equal to some value, then the resulting equation can be solved for one or more values of y. If we specify which value of y to take, we have a function. In this case we call y an implicit function of x. If we wanted to find the slope to the curve x 2 + y 2 = 1 at a certain point, we could solve for y to obtain an explicit function of x. We would yield y = 1 x 2 or y = 1 x 2, and as long as the point we care about isn t (1, 0) or ( 1, 0), we could then take the derivative of the appropriate choice for y to find the slope of the tangent line. This has two issues. (i) We cannot always write y as an algebraic expression in terms of x. (ii) Even if we can solve for y, what we obtain might be very complicated. Instead we will use a technique known as implicit differentiation. This technique involves assuming that y is a function of x, say y = f(x) for some (possibly non-algebraic) function f. From the given equation x 2 + y 2 = 1, we get the equation x 2 + [f(x)] 2 = 1. We view both sides of the above equation as a function of x, say g(x) = x 2 + [f(x)] 2 and h(x) = 1. Since g(x) = h(x), then we should have g (x) = h (x). Clearly h (x) = 0, so we have 0 = g (x) = d ( x 2 + [f(x)] 2) = 2x + d dx dx [f(x)]2 = 2x + 2f(x)f (x). We can then solve for f (x) in this equation to obtain f (x) = x/f(x) = x/y. Now if we wanted to find the slope at a point (a, b) on the graph of the equation x 2 + y 2 = 1, we get f (a) = a/b. We didn t ever have to actually know what f(x) was! 1
Notice we assumed that y = f(x) for some function f. There is a theorem called the implicit function theorem that gives specific conditions under which this assumption is guaranteed. We won t worry about this. Also, notice it was easy to solve for f (x). This is because the resulting equation was linear in f (x). Thankfully, this is always the case. This makes implicit differentiation much easier than the (potentially) more complicated method of solving for y before differentiating. Let s review the steps we used for implicit differentation. Steps for implicit differentiation Start with an equation involving x and y. We assume y is a function of x. We want to find dy dx. (a) Apply d dx (b) Solve for dy dx. Examples d dy to both sides of the equation. Remember that (y) = dx dx d dx (g(y)) = g (y) dy dx. and, by the chain rule, 1. Let x 4 2xy 3 + y 5 = 32. Find y and the slope of the tangent line to the curve when x = 0. 2
2. The demand function for a certain commodity is given by p = 13 1 + x 2 + x 3. Find the instantaneous rate of change of the number sold with respect to the price when x = 2. 3
Related rates One consequence of the chain rule is that the rate of change of one quantity is related to the rate of change of another quantity. This comes into play when each of our quantities are changing with respect to time. Related rates is just what you would think. It is the study of problems in which different rates are related in some way. It s best to see how this works through an example. Suppose you are blowing up a spherical baloon by inputting 300 cubic inches of air per minute, and you want to know the rate at which the radius is increasing. We start with the relationship between the radius r and the volume V of the baloon, which is V = 4 3 πr3. Now, viewing both V and r as a function of time t, we implicitly differentiate the above equation with respect to t. This yields the equation dv = 4 dr π3r2 3 = 4πr2 dr. Since we are interested in the rate at which the radius is changing, we solve for dr dr = dv 4πr 2. to obtain Now we can determine the rate of change of the radius. For example, when the baloon has a radius of 1 inch, the radius is increasing by 300 23.87 inches per minute. 4π(1) 2 When the baloon has a radius of 3 inches, the radius is increasing by 300 2.65 inches per minute. 4π(3) 2 Many problems dealing with related rates will require creating a mathematical model. Thus you shouldn t forget the guidelines from section 5.6. 4
3. Suppose the cost function in thousands of dollars is given by C(x) = 100 + x 2, where x is in thousands of units. The company estimates that its rate of change of sales with respect to time will be 0.04. What will be the rate of increase of costs with respect to time if the company produces and sales 10, 000 items? 4. An oil slick is spreading in a circular fashion. The radius of the circular slick is observed to be increasing 0.1 mile per day. What is the increase of the area of the click when r = 2 miles? 5
5. Two ships leave the same port at 1:00 PM. The first ship heads due north, and at 2:00 PM is observed to be 3 miles due north of port and going at 6 miles per hour at that instant. The second ship leaves port at the same time as the first ship, and heads due east, and 1 hour later is observed to be 4 miles due east of port and going 7 miles per hour. At what rate is the distance between the two ships changing at 2:00 PM? Exercises 1. Differentiate implicitly and find the slope of the curve at the indicated point. (a) xy + x + y 2 = 7, (1, 2) (b) x 2 y + x y 3 = 1, (2, 1) 2. Find dy given the indicated information. (a) y = 2 + x 2, dx (b) y = 1 x 4, dx = 4, x = 2. = 1, x = 2. (c) 3x 2 y 2 = 23, (3, 2) (d) x 3 2y 3 = 15, (1, 2) (c) x 2 + y 2 = 5, dx = 2, x = 1, y = 2. (d) y = 1 x 1 + x, dx = 2, x = 1. 2 3. After using an experimental drug, the radius of a spherical tumor is observed to be decreasing at a rate of 0.1 cm per week when the radius is 2 cm. At what rate is the volume of the tumor decreasing? [Hint: Volume of a sphere of radius r is (4/3)πr 3.] 4. Do exercises 5.8.41-5.8.51 from the textbook. 6