Uses for the Derivative

Applications of the derivative.

For most applications, we think of f¢(x) as measuring the rate of change of the function f with respect to change of the variable x . That is one reason why we denote the derivative sometimes by df/dx , thinking of it as the ratio. For functions of time (usually written as t ) the idea of rate makes the most sense, but we can also talk about the rate of change of a function with respect to, say, distance as well.

Here are some standard examples of how the derivative is interpreted.

2.4.1  Position and velocity:

If f(t) measures the position of a particle (or whatever) as it moves along a line, then the derivative of the position, f¢(t) = df/dx : = v(t) is its velocity, and the derivative of the velocity, v¢(t) = dv/dt : = a(t) is the acceleration of the particle.

Example 1 If the position of a particle (in meters, along a line) is given by s = t3-4.5t2-7t , for t ³ 0 , find the velocity v(t) , and find when the velocity reaches 5 m/sec . What motion did the particle have at t = 0 ?

Example 2 A ball is thrown vertically upward with a velocity of 80 ft/sec , and its position t seconds later is s(t) : = 80t-16t2 feet up. How high does it get, and how fast is it moving when it hits the ground again?

Exercise 1 A ball is thrown vertically upward with a velocity of 80 ft/sec , and its position t seconds later is s(t) : = 80t-16t2 feet up. How high does it get, and how fast is it moving when it hits the ground again?

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2.4.2  Density

The density r of a bar or rod is the mass per unit length. But if the bar is not uniform (perhaps not uniform thickness), the density will not be constant, so to really have a good idea of the density of the rod at a given point x along its length, we have to take the infinitesimal ratio of mass per length at x . If you think of the mass of the rod up to the point x as m(x) , then the density is the rate of change of m(x) , r = dm/dx .

Later on, in other courses, you'll get more interpretations of this notion of density. Clearly, really, density should be mass per unit volume, but that requires, in its full generality, techniques of third-semester calculus. So, for the time being, we'll stick to a one-dimensional model.

Example 3 Assume that the thickness of a wire increases slightly along the length, so that the mass up to x meters is m(x) : = 3x+x2/10 grams. Find the density three inches from the beginning of the wire.

Solution:

r(x) = dm/dx = 3+x/5 , so r(3) = 3.6 g/m .

2.4.3  Cost

The total cost of producing x items is called (surprisingly) the cost function, C(x) . The derivative, C¢(x) , is the marginal cost, and represents (approximately) the cost of producing the next item. The word ``marginal'' in economics refers to a rate of change, so that your ``marginal tax rate'' is the tax percentage on the next dollar you earn.

Typically, the costs involved in producing x items include a certain amount of fixed costs, which would include the cost of the factory, machinery, and other things you need no matter how many of the widgets you produce. In addition, materials and labor costs typically are proportional to the number of items produced, but there might be some costs that grow more quickly as the number of items increase, such as warehousing costs. So, a typical cost function might look like C(x) = 300+20x+x2/10 .

Example 4 Presume that the cost function to produce x widgets is C(x) = 300+20x+x2/10 . Then, find the marginal cost at the production level of 100 widgets.

Solution:

The marginal cost is C¢(x) = 20+x/5 , so the marginal cost at a production level of 100 widgets is C¢(100) = 40 .

2.4.4  Linear approximation

The Leibniz notation for derivatives, df/dx , isn't supposed to mean a fraction, really. But, in reality it does have some meaning like that. Since the derivative is the limit of fractions, if h is near a , then (f(a+h)-f(a))/h will be near f¢(a) . That is, replacing a+h by the more generic x and h by x-a :
f(x)-f(a)
x-a
 » f¢(a),

or, solving for f(x) ,
f(x) » f(a)+f¢(a)(x-a).

Another way to look at this is that the equation of the tangent line L to the curve y = f(x) at (a,f(a)) is y = f(a)+f¢(a)(x-a) . Since the tangent line is near the graph, the y will be almost the same as f(x) .

The point here is to approximate f(x) , which might be a very complicated function, by a simpler function - a linear one, which is why it is called linear approximation.

We sometimes write Dx for x-a (or h ), to indicate the change in x . In that notation:
f(a+Dx) » f(a)+f¢(a)Dx,
or writing Df: = f(a+Dx)-f(a) to denote the change in f (the D usually refers to change in the quantity), Df » f¢(a)Dx. We sometimes call the right-hand side of that equation the differential of f , but when we do we usually write dx instead of Dx . That is,
df = f¢(a)dx,
getting us back to the Leibniz notation for f¢(x) , df/dx . The point of that is to approximate the function by the linearization, the differential.

Example 5 Find an approximate value of
  ___
Ö4.2
 
  

Example 6 Use differentials to find the relative error (the percentage of error) in the computation of the volume of a cube, if the measurements of the sides might be off by as much as 1%.

Exercise 2 Approximate
  ___
Ö8.7
 
using differentials (equivalently, using linear approximation).

Answer

Exercise 3 Find the differential of x3+2x .

Answer

Exercise 4 Find the linear approximation of
  ___
Ö1+x
 
for x near 0.

Answer

Exercise 5 Find the linear approximation of

Ö
 
1+x2
 
for x near 0.

Answer

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2.4.5  Related Rates Problems

Related Rates problems are one of the classic types of ``word problems'' in calculus classes. They provide a first introduction to mathematical modeling: taking a problem, distilling from it the mathematical essentials, then, using standard mathematical techniques which can be applied to any problem of the same general type, solve the resulting mathematical problem.

For the problems we'll be looking at now, the underlying mathematics is quite simple. Let's say we have two functions, f and g . We can make them both functions of t to keep in the spirit of many of these applications. Then, if we know a relationship between f and g (some equation involving them), we can differentiate both sides of the equation to get a relationship between their derivatives. In applications, we usually know something about one of the derivatives, and we want to figure out the other.

For example, if
(f(t))2+(g(t))2 = 1,
then differentiate both sides. The derivative of 1 is 0, and the derivative of (f(t))2 (with respect to t ) is 2f(t)f¢(t) . Similarly, ((g(t))2)¢ = 2g(t)g¢(t) , so that
2f(t)f¢(t)+2g(t)g¢(t) = 0,

an equation involving the derivatives of f and g (the rates of the functions).

Here is how the idea works in practice

Example 7 A boater has his boat tied to the pier by a long rope. He feeds the pier-end of the rope through a winch and pulls his boat in towards the pier. The tide is out, so the bow of the boat is 6 feet below the level of the pier. If the winch is pulling in the line at a constant rate of 2 feet per second, how fast is the boat moving towards the pier when there is 10 feet of rope between the boat and the winch?

 

Exercise 6 Egbert is flying his kite. It is 75 feet off the ground, moving horizontally away from Egbert at a rate of 5 feet per second. How fast is Egbert letting out the string when the kite is 100 feet (horizontally) downwind of him?

Answer

Exercise 7 Harpo and Chico are walking away from Groucho, in paths perpendicular to each other. Harpo is 4 feet from Groucho, walking at 3 feet per second, and Chico is 3 feet from Groucho, walking at 5 feet per second. How fast is the distance between Harpo and Chico changing?

Remark If you're curious about who these characters are, take a look at this link Marx Brothers to one of their fan sites. They were called the Marx Brothers, and starred in several hilarious movies in the thirties and forties.

Answer

Example 8 A man, walking at night, is walking directly towards a streetlight. The light is 10 feet off the ground, and the man is 6 feet tall and walking at 4 feet per second. When the man is 8 feet from the streetlight, how fast is the length of his shadow changing?


Exercise 8 A 20 foot ladder is leaning against a wall. A painter stands on the top of the ladder, minding his own business. Some fool comes by and ties his dog to the base of the ladder, a cat comes along, and the dog chases after the cat, dragging the base of the ladder with him at a rate of 2 feet per second directly away from the wall. How fast is the painter falling when he is 12 feet from the ground?

Answer

Exercise 9 Egbert is drinking a daquiri (non-alcoholic) out of a glass that is a cone with a radius at the top of 3 inches and a height of 5 inches. He drinks his daquiri at a constant rate of 2 cubic inches per second (through a straw). How fast is the top of the daquiri falling when it is 4 inches above the bottom of the glass?

Answer

Exercise 10 A baseball diamond is in the shape of a square, with sides of length 90 feet. Who's on first base, What's on second, and Eyedonno's on third. When the ball is hit, What runs towards Eyedonno on third, and Who runs towards second. When What is 60 feet from Eyedonno (still on third base), running towards third at 20 feet/sec, and Who is 40 feet from second, running towards second at 25 feet/sec, how fast is the distance between Who and What changing?

Note 1 If you sort of recognize what this is all about, then you win a prize. The plot of this problem is shamelessly stolen from one of the funniest stand-up routines of all time, the Who's on First? routine of Abbott and Costello. Read about it here. The real routine is a lot funnier than this problem.

Answer

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Copyright 2000 David L. Johnson