How Do You Undo A Rate with the Substitution Method?

Suppose that the profit for a company is increasing at a rate of

undo_rate_01

where the company has been in operation for t years. What is the total change in profit over the first three years?

In this problem, we are given the rate at which profit is changing over time. This is confirmed by the fact that the function is defined as P′(t), the derivative of profit. However, the question is about the corresponding profit function P(t). So we need to find this profit function by taking the antiderivative of P′(t),

undo_rate_02

Remember, the antiderivative undoes the derivative so the antiderivative of P′(t) is P(t). To do this antiderivative, we need to use the Substitution Method.

undo_rate_03

This means that

undo_rate_04

You might think that the total change in profit over the first three years is P(3), but this is the profit at the end of the third year. To find the total change in profit we need to calculate P(3) – P(0),

undo_rate_05

 

How Do I Find Area Between Curves?

Sometimes a seemingly easy problem can get fairly complicated with the addition of a few extra requirements. For instance, suppose we want to find the area between the functions y = x2 – 4 and y = 3x. If we graph the two functions, we see that they appear to cross at x = -1 and x = 4.

area_btwn_curves_01

We can verify these two points by setting the functions equal to each other and solving for x.

area_btwn_curves_06

The area between these curves lies above the parabola and below the line.

area_btwn_curves_02

To find the area of the shaded region, take the definite integral from x = -1 to x = 4 of the higher function minus the lower function,

area_btwn_curves_07

We can evaluate this integrand using the Fundamental Theorem of Calculus,

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Let’s now modify this problem by finding the area enclosed by the functions y = x2 – 4 and y = 3x as well as the lines x = -5 and x = 1. Graph each of these equations.

area_btwn_curves_03

The region enclosed by these graphs is more complicated since the functions cross at x = -1.

area_btwn_curves_04

To find the area of the enclosed region, we need to break it into two parts. The first part runs from x = -5 to the point of intersection at x = -1. The area of this part is

area_btwn_curves_09

The second part extends from x = -1 to x = 1 and has area

area_btwn_curves_10

The area enclosed by the functions y = x2 – 4 and y = 3x as well as the lines x = -5 and x = 1 is the sum of these parts or 206/3. Adding the vertical lines on either side of the point of intersection requires the use of two definite integrals since the parabola is higher on the left side of x = -1 and the line is higher on the right side of x = -1.

area_btwn_curves_05

Thus the area between the curves is 184/3 + 22/3 or 206/3.

How Do You Calculate Producers’ and Consumers’ Surplus?

In Section 14.3, I carry out several examples where the producers’ or consumers’ surplus is calculated. I want to give you a few more examples including some of the examples worked out by students in class.

Let’s take a look at producers’ surplus. To get a good idea of this concept, let’s visualize what area on a supply or demand graph represents. In the graph below, we have a supply function displaystyle S(Q)=0.9Q. The supply and demand are in equilibrium when 100 units are produced at 90 dollars per unit.

On this graph heights are in dollars per unit and widths are in units. This means the units on any area will be

displaystyle frac{text{dollars}}{text{unit}}cdot text{units}=text{dollars}

The area Thunder the supply curve is

This is the amount of money a supplier would be willing to receive if each of the units yielded revenue according to the prices on the supply curve from 0 to 100 units.

However, if the market is in equilibrium all 100 units with earn 90 dollars per unit yielding

 

Since the market is in equilibrium, the supplier actually receives a higher price per units giving an additional 4500 dollars in revenue. This extra amount is called the producers’ surplus.

The consumer would be willing to pay more than the equilibrium price. The amount they save by paying the equilibrium price is called the consumers’ surplus.

Problem 1 Suppose the supply curve for a particular product is

displaystyle S(Q)={{Q}^{{scriptstyle{}^{5}!!diagup!!{}_{2};}}}+2{{Q}^{{scriptstyle{}^{3}!!diagup!!{}_{2};}}}+50

and that the equilibrium quantity is Q = 16. Find the producers’ surplus.

First find the equilibrium price (black). Then find the area under the supply curve (red) and the area under the equilibrium price (green). The difference between these amounts (blue) is the producers’ surplus.

Notice that the producers’ surplus is the area between the equilibrium price and the the supply curve. We can compute the surplus by computing this area.

This gives rise to the formula often quoted for the  producers’ surplus,

displaystyle text{Producers } text{Surplus}=intlimits_{0}^{{{Q}_{e}}}{ left( {{P}_{e}}-Sleft( Q right) right)} dQ

A similar formula exists for the consumers’ surplus and is essentially the area between the demand curve and the equilibrium price,

displaystyle text{Consumers } text{Surplus}=intlimits_{0}^{{{Q}_{e}}}{ left( D(Q)-{{P}_{e}} right) dQ}

Problem 2 Suppose the demand and supply curves for a product are

displaystyle D(Q)=900-20Q-{{Q}^{2}}

displaystyle S(Q)={{Q}^{2}}+10Q

Find the producers’ and consumers’ surplus.

Start by finding the equilibrium point (black). Then find the producers’ surplus (blue).


The consumers’ surplus is

How Do You Use the Substitution Method to Calculate Change?

Only one problem on the Section 14.1 Homework was missed by very many people. In that problem you were given the rate of change of profit, P‘(t), and asked to calculate how much the profit changed. Since this is a question about P(t), you need to undo the derivative with an antiderivative in the form of the Fundamental Theorem of Calculus. With this function, we would write it as

displaystyle intlimits_{a}^{b}{P'(t),dx=P(b)-P(a)}

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