Calculus, by Smith and Minton

2.8 Implicit Differentiation and Related rates

Compare the following two equations describing familiar curves:

y = x²+3 (parabola)

and

x²+y² = 4 (circle).

The first equation defines y as a function of x explicitly, since for each x, the equation gives an explicit formula y = f (x) for finding the corresponding value of y. On the other hand, the second equation does not define a function, since the circle in Figure 2.40 doesn't pass the vertical line test. However, you can solve for y and find two functions (y = and that are defined implicitly by the equation x²+y² = 4. Graphically, we identify these implicit functions as portions of the curve defined by the original equation, each one of which must pass the vertical line test. We need to be able to compute slopes of tangent lines to these curves even if we can't solve the defining equation explicitly for y.

Suppose that we want to find the slope of the tangent line to the circle x²+y² = 4 at the point (1, -) (see Figure 2.40).

Figure 2.40
The tangent line at the point (1, -).

Our derivative formulas are not directly applicable because we do not have y written explicitly as a function of x. We can, of course, think of the circle as the graph of two semicircles,

and

Since we are interested in the point (1, -

), we use the equation describing the bottom semicircle,

This is now an explicit formula for y in terms of x and we can use this to compute the derivative

So, the slope of the tangent line at the point (1, -

) is then

This calculation was not especially challenging, although we will soon see an easier way to do it. However, it's not always possible to explicitly solve for a function defined implicitly by a given equation. For example, van der Waal's equation relating the pressure P, volume V and temperature T of a gas has the form

(8.1)

where a, n, b and R are constants. Notice the difficulty in solving this for V as a function of P. If we want the derivative

we will need a method for computing the derivative directly from the implicit representation given in (8.1).

Consider each of the following calculations:

and

Notice that each of these calculations has the form

for some choice of the function y(x). This last equation is simply an expression of the chain rule. We can use this notion to find the derivatives of functions defined implicitly by equations.

We first return to the simple case of the circle x²+y² = 4. Assuming this equation defines one or more functions of x: y = y(x), the equation is

(8.2)

Differentiating both sides of equation (8.2) with respect to x, we obtain

Using the general rule

we get

2x+2y(x)y' (x) = 0.

(8.3)

We can solve equation (8.3) for the derivative y' (x), as follows. Subtracting 2x from both sides gives us

2y(x)y' (x) = -2x

and dividing by 2y(x) (assuming this is not zero), we get

Notice that here y'(x) is expressed in terms of both x and y. To get the slope at the point (1, -

), substitute x = 1 and y = -

. We have

(Notice that this is the same as we had found earlier by first solving for y explicitly and then differentiating.) This process of differentiating both sides of an equation with respect to x and then solving for y'(x) is called implicit differentiation.

When faced with an equation implicitly defining one or more functions y = y(x), differentiate both sides with respect to x, taking care to use the chain rule wherever applicable. In particular, be careful to recognize that differentiating any function of y will require the chain rule:

Then, gather any terms with a factor of y'(x) on one side of the equation, with the remaining terms on the other side of the equation and solve for y' (x). We illustrate this process in the next several examples.

8.1

Finding Slope Implicitly

Find y'(x) for x²+y³-2y = 3. Then, find the slope of the tangent line at the point (2, 1).

Since we can't (easily) solve for y in terms of x explicitly, we compute the derivative implicitly. Differentiating both sides with respect to x, we get

and so,

2x+3y²y'(x)-2y'(x) = 0.

To solve for y' (x), simply write all terms involving y'(x) on one side of the equation and all other terms on the other side. We have

3y²y'(x)-2y'(x) = -2x

Subtracting 2x from both sides

and hence, after factoring, we have

(3y²-2)y'(x) = -2x.

Factoring y'(x) from both terms on the left side.

Solving for y'(x), we get

Dividing by (3y² - 2).

Substituting x = 2 and y = 1, we find that the slope of the tangent line at the point (2, 1) is

The equation of the tangent line is then found from

y-1 = -4(x-2).

We have plotted a graph of the equation and the tangent line in Figure 2.41 using the implicit plot mode of our computer algebra system.

Figure 2.41
Tangent line at (2, 1).

8.2

Finding a Tangent Line by Implicit Differentiation

Find y'(x) for x²y²-2x = 4-4y. Then, find an equation of the tangent line at the point (2, -2).

Differentiating both sides with respect to x, we get

Since the first term is the product of x² and y², we must use the product rule. We get

2xy²+x²(2y) y'(x)-2 = 0-4y'(x).

Grouping the terms with y' (x) on one side and terms without y' (x) on the other, we get

(2x²y+4)y' (x) = 2-2xy²

and we can solve for y' (x):

Substituting x = 2 and y = -2, we get the slope of the tangent line,

Finally, an equation of the tangent line is given by

y+2 =

(x-2).

We have plotted the curve and the tangent line at (2, -2) in Figure 2.42 using the implicit plot mode of our computer algebra system.

Figure 2.42
Tangent line at (2, - 2).

You can use implicit differentiation to find a needed derivative from virtually any equation you can write down. We illustrate this next for an application.

8.3

Rate of Change of Volume with Respect to Pressure

Suppose that van der Waal's equation for a specific gas is

(8.4)

Thinking of the volume V as a function of pressure P, use implicit differentiation to find the derivative

at the point (5, 1).

Differentiating both sides of (8.4) with respect to P, we have

From the product rule and the chain rule, we get

Separating the terms that contain a factor of

from those that do not, we get

Solving for

we get

and so,

(The units are in terms of volume per unit pressure.) We show a graph of van der Waal's equation, along with the tangent line to the graph at the point (5, 1) in Figure 2.43.

Figure 2.43
Graph of van der Waal's equation and the tangent line at the point(5,1).

Recall that, up to this point, we have proved the power rule

only for integer exponents (see Theorems 3.3 and 4.3), although we have been freely using this result for any real exponent, r. Now that we have developed implicit differentiation, however, we finally have the tools we need to prove the power rule for the case of any rational exponent.

8.1

For any rational exponent, r,

Suppose that r is any rational number. Then

for some integers p and q. Let

y = x^r = x^p/q.

(8.5)

Then, raising both sides of equation (8.5) to the qth power, we get

y^q = x^p.

(8.6)

If we differentiate both sides of equation (8.6) with respect to x, we get

From the chain rule, we have

It's now a simple matter to solve for

We have

		Since y = x^p/q.
		Using the usual rules of exponents.
		Since

as desired.

Related Rates

As we have mentioned, you can use implicit differentiation to find a derivative from nearly any equation. In many situations, each quantity in the equation changes with time (or some other variable). In this case, any derivative we find is called a related rate, since we can see from the differentiated equation how the derivatives (rates) are related. We illustrate this idea in the next several examples.

8.4

A Related Rates Problem

An oil tanker has an accident and oil pours out at the rate of 150 gallons per minute. Suppose that the oil spreads onto the water in a circle at a thickness of

'' (see Figure 2.44). Given that 1 ft³ equals 7.5 gallons, determine the rate at which the radius of the spill is increasing when the radius reaches 50 feet.

Figure 2.44
Oil spill.

Since the area of a circle of radius r is

r² , the volume of oil is given by

V = (depth) (area) =

r²,

since the depth is

'' =

ft. Both volume and radius are functions of time, so

Differentiating both sides of the equation with respect to t, we get

We are given a radius of 50 feet. The volume increases at a rate of 150 gallons per minute, or

/min. Substituting in V'(t ) = 20 and r = 50, we have

Finally, solving for r' (t ), we find that the radius is increasing at the rate of

7.6394 feet per minute.

Although the details change from problem to problem, the general pattern of solutions is the same for all related rates problems. Looking back, you should be able to identify each of the following steps in example 8.4.

1. Make a simple sketch, if appropriate.
2. Set up an equation relating all of the relevant quantities.
3. Differentiate both sides of the equation.
4. Substitute in values for all known quantities and derivatives.
5. Solve for the remaining rate.

8.5

Another Related Rates Problem

A car is traveling at 50 mph due south at a point

mile north of an intersection. A police car is traveling at 40 mph due west at a point

mile due east of the same intersection. At that instant, the radar in the police car measures the rate at which the distance between the two cars is changing. What does the radar gun register?

First, we sketch a picture and name the vertical distance of the first car y and the horizontal distance of the police car x (see Figure 2.45).

Figure 2.45
Cars approaching an intersection.

Notice that this says that

since the police car is moving in the direction of the negative x - axis and

since the other car is moving in the direction of the negative y - axis. From the Pythagorean Theorem, the distance between the two cars is

Since all quantities are changing with time, our equation is

Differentiating both sides with respect to t, we have by the chain rule that

d' (t )	= {[x(t )]²+[y(t )]²}^-1/22[x(t ) x' (t ) +y(t )y' (t )]

Substituting in x(t ) =

, x'(t ) = -40, y(t ) =

and y' (t ) = -50, we have

so that the radar gun registers 62.6 mph. Note that this is a poor estimate of the car's actual speed. For this reason, police nearby always take radar measurements from a stationary position.