Slope of a Curve
A chord is a straight line which cuts the curve at two points.
A tangent is a straight line which touches the curve at one point only.
The slope or gradient, of a curve at a point P is the slope
of the tangent to the curve at that point P.
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Slope varies as one moves along a non-linear curve.
| At | A, | slope is negative
| | | B, | slope is zero
| | | C, | slope is positive
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The slope of (the graph) of a function is the derivative of the function.
If the function is given as a
rule
f(x), the derivative is denoted by
f'(x). If we write
y = f(x) then
f'(x) is also denoted by
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The slope of the function at P is got by evaluating the slope of the cords PQ1,
PQ2 and letting the point Q approach P. The limiting value gives the slope of the tangent at P
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Suppose
P has coordinates
(a, f(a))
and that
Q has coordinates
(a + Δx, f(a + Δx))
| Slope of chord PQ |
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Now let
Q tend to
P. Then
Δx -> 0,
Slope of
PQ -> slope of tangent to curve at
P, and
But, slope of tangent to curve at
a is
f'(a) so we get,
Example. Let
f(x) = 2x2 + 1. Find the slope of the graph of
f(x) when
x = 2
We need to evaluated
f'(2), which is
If
f(x) = 2x2 + 1, then
This last expression tends to 8 as
Δx -> 0,
so we get
Fortunately, we do not need to go through this type of calculation each time we need a derivative; we use rules.
Rules give the derivative function or the derived function; we then insert the appropriate
x
value to get the derivative (or slope) at a particular point.
Higher Derivatives
Turning Points
Maximum and Minimum Turning Points
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| (local) minimum | (local) maximum
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Slope goes from negative, to zero, to positive so | Slope goes from positive, to zero, to negative so
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| Read Worked | Example 6.18 ; pages 314, 317
| (Example 6.17; pages 266, 267 of 2nd Ed.) | |
Example (#6, page 316; page 267 of 2nd Ed.) Find the turning points for the curve
Determine which are local maximum and which are local minimum points.
Solution
| Turning points are where |  |
Turning points when
x = 3 and
x = -1
| At x = -1, |  |
Thus x = -1 is a local maximum
| At x = 3, |  |
Thus x = 3 is a local minimum
Coordinates of these points are (-1, 5) and (3, -27)
Sketch of graph.
- Points where curve cuts coordinate axes
- Maximum and minimum points
- Intervals where curve increases or decreases
Example (#10, page 325; page 276 of 2nd Ed.) Sketch the curve
Solution
Step 1. Find derivatives
Step 2. Find turning points
are turning points
Step 2a. Find
y-coordinates at each
T.P.
Step 3. Nature of turning points
Step 4. Intercepts with coordinate axes
Step 5. Mark above points and plot.
| Read Worked | Example 6.20 ; pages 318-320
| (Example 6.19; pages 269-272 of 2nd Ed.) | |
- Exercises;
| # 1, 2, 6, 7, 9 | page 325
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| (page 276 of 2nd Ed) |
The first derivatives of economic functions are called marginal functions.
Therefore, the optimum value of functions,
such as revenue, profit, cost etc, will occur when
the corresponding marginal function (first derivative) is zero.
Example (#1, page 340; page 290 of 2nd Ed.) A firm's total revenue is given by the equation
| a) | | Find the value of Q for which TR is maximized. Hence calculate the maximum TR
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| b) | | Write down the equations of the average revenue and marginal revenue functions. Describe how AR and MR
change before and after the maximum TR
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| c) | | Sketch TR, MR and AR on the same diagram. From the graph, write down the values of AR and MR when TR is zero and when TR is a maximum.
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Solution
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| | | If MR = 0, Q = 15 and TR = 675
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| | | | | For Q < 15 ; | AR and MR are positive but
| | | decreasing. At Q = 15, MR = 0
| | For Q > 15 ; | MR < 0 while AR is positive until
| | | Q = 30
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| c) | | When TR = 0, AR = 90 and MR = 90
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| | | When TR is a maximum (i.e Q = 15)
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Example (#3, page 341; page 290 of 2nd Ed.) A shop which sells T-shirts has a demand function and a total cost function given by,
| a) | | Write down the equation for TR and profit.
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| b) | | Calculate the number of T-shirts which must be sold to maximize i) profit, ii) total revenue.
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| c) | | Write down the equations for MR and MC. Show that MR = MC when profit is maximized.
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| d) | | Plot the graphs of
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| | | i) TR and TC on the same diagram. From the graph estimate the break-even points. Conform your answer algebraically.
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| | | ii) MR and MC on the same diagram. What is the significance of the point of intersection of these two graphs?
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Solution
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| d) | | i)
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| | | ii) Exercise
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| Read Worked | Examples 6.22, 6.23, 6.25; pages 325-335
| (Examples 6.21, 6.22, 6,24; pages 277-285 of 2nd Ed.) | |
- Exercises;
| # 2, 4, 5 | pages 340, 341
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| (pages 290, 291 of 2nd Ed) |
Further Differentiation
Composite functions.
To differentiate, use the chain rule.
| Write first function as |
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In above examples;
The derivative is
| So in case of |  |
Example Differentiate the following
Solution
Product Rule
If
u is a function of
x and
v is a second function of
x
Example Differentiate the following.
Solution
Quotient Rule
| If |  | , then |
Example Differentiate the following.
Solution
Summary
Rules Used
- Power Rule
- Exponential Rule
- loge = ln Rule
- Chain Rule
- Product Rule
- Quotient Rule
Example (#4, page 376; page 319 of 2nd Ed.) A firm's average revenue function is
| a) | | Determine the equation for TR and MR
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| b) | | find the value of Q for which TR is a maximum. Calculate the price and TR when TR is a maximum.
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Solution
| a) | |
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| b) | | TR is a maximum when MR = 0, i.e
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| | | When TR is a maximum, i.e Q = 20, then
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| | | Price is average revenue so at Q = 20
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Example (#6, page 376; page 319 of 2nd Ed.) Given the average cost function
| a) | | Find the minimum AC and graph the AC from Q = 0 to Q = 25.
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| b) | | Write down the equation of TC
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Solution
Point Elasticity of Demand
Price elasticity of demand.
Point elasticity of demand is then given in terms of
dQ/dP as,
Example (#3, p386; page 326 of 2nd Ed.) The demand for family membership of a sports club is given by the equation,
where
P is the monthly fee.
| a) | | Derive an expression for the price elasticity of demand in terms of Q. Evaluate
εd when Q < 100 and when Q > 100.
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| b) | | If the club has 150 members calculate the fee per membership (price). If the fee increases by 10%,
use the elasticity to calculate the approximate % change in demand.
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Solution
| a) | |
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| | | If Q < 100, εd < -1. This means that % change in Q
is opposite in direction to that of P and is greater in size since.
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| | | If Q > 100, % change in Q is again in
opposite direction but less in size than change in prize %.
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| Read Worked Examples 6.40, and 6.43 | ; pages 380, 384-386
| (pages 322-324 of 2nd Ed.) | |
- Exercises;
| # 4, 9 | pages 386, 387
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| (pages 326, 327 of 2nd Ed) |
Example Given the demand function
find the price elasticity of demand when
P = 51. If this price rises by 2% calculate the corresponding
% change in demand.
Solution
We need to evaluate
εd when
P = 51, so
first get corresponding value for Q and then substitute into (1) to get
εd.
If
P, solve for
Q as follows,
Ignore
Q = -9. So when
P = 51,
Q = 5 and from (1)
Review Problems
Example Evaluate the derivative of the following function at
x = 2.
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Solution
Example
| a) | | The fixed cost of producing a product are 100 and the variable costs are 2 + Q/10 per unit.
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| | i) | Find an expression for total costs (TC) and marginal costs (MC)
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| | ii) | At what level of output does MC = 22?
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| b) | | If the supply equation is
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Find the price elasticity of supply if the current price is 80. Estimate the percentage change in supply if the price increases by 5%.
Solution
Example If
f(x) = x3 - 3x2 + 2x, find the intervals where
f(x) is increasing or decreasing and the coordinates of the maximum or minimum point. Sketch the graph of
f(x)