Unit 4 - 2

Quick Derivative Review

Example: On the graph of y = 2 sin(5x) + 3, what is the slope at x = π/4?

Example: On the graph of x = 14e−t − 3t, at what rate is x changing att = 2?

Unit 4 - 3

Interpreting the Derivative

Example: Consider the statement “I am walking at 1.2 m/s.”How far will you travel in the next second?

How far will you travel in the two seconds?

How far will you travel in the next1

3of a second?

How far will you travel in the next 10 minutes?

Unit 4 - 4

Note that all the values computed above are estimates or predictions. Whichof the estimates you just calculated will be the most accurate?

What assumptions are you using to reach your answers?

Unit 4 - 5

Example: Let R = f(A) be the monthly revenue for a company, given ad-vertising spending of A per month. Both variables are measured in thousandsof dollars.

Interpret f ′(200) = 1.8 in words.

Unit 4 - 6

If A = 200 currently, and you increased advertising spending by 2 thousanddollars, what would you expect your revenue increase to be?

If A = 200 currently, and you increased advertising spending by 1 milliondollars, what would you expect your revenue increase to be?

If f ′(200) = 0.8, and you are currently spending 200 thousand on advertising,should you spend more or less next month?

Unit 4 - 7

Question: A chemical reaction consumes reactant at a rate given by f(c), wherec is the amount (mg) of catalyst present. f(c) is given in moles per second.The units of the derivative, f ′(c), are

(a) mg/s

(b) moles/s

(c) moles/(s mg)

(d) (mg moles)/s

Unit 4 - 8

Question: If f ′(10) = −0.2,

(a) Adding more catalyst to the 10 mg present will speed up the reaction.

(b) Adding more catalyst to the 10 mg present will slow down the reaction.

(c) Removing catalyst, from 10 mg present, will speed up the reaction.

(d) Removing catalyst, from 10 mg present, will slow down the reaction.

Unit 4 - 9

Local Linearity

In all these estimates we have been making, we have been relying on the locallinearity of a differentiable function.

If a function is differentiable at a point, then it behaves like a linear function for xsufficiently close to that point.

Another interpretation of differentiability is that if we “zoom in” sufficiently on apoint, the graph will eventually look like a straight line.

Unit 4 - 10

Consider the graph of y = sin(x) at different scales, around the point x = 0.4:

−8 −6 −4 −2 0 2 4 6 8 10

−6

−4

−2

0

2

4

6

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−1.5

−1

−0.5

0

0.5

1

1.5

−0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5

0.3

0.32

0.34

0.36

0.38

0.4

0.42

0.44

0.46

Unit 4 - 11

And the more exotic y = sin(1/x) at different scales, around x = 0.1:

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

−1.5

−1

−0.5

0

0.5

1

1.5

−0.05 0 0.05 0.1 0.15 0.2 0.25−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0.09 0.095 0.1 0.105 0.11−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.099 0.0992 0.0994 0.0996 0.0998 0.1 0.1002 0.1004 0.1006 0.1008 0.101−0.64

−0.62

−0.6

−0.58

−0.56

−0.54

−0.52

−0.5

−0.48

−0.46

−0.44

Unit 4 - 12

In both cases, the function is differentiable at the point we’re considering, and (asa consequence), the graph looks linear from a sufficiently close viewpoint.

Unit 4 - 13

Sketch a graph of a locally linear function f(x). Add on the tangent line, anduse the derivative to estimate ∆y for a given change in x.

Unit 4 - 14

Derivative as Approximation of Change

f ′(x) =dy

dx≈

∆y

∆xso given a value of ∆x,

∆y ≈ f ′(x)∆x

assuming that ∆x is “sufficiently” small.The larger ∆x is, the worse the approximation will generally be.

Unit 4 - 15

Tangent Line or Local Linearization

In the last few examples, we focused on the change in y (or f , or revenue, etc.),based on a set change in the input. Note that all these changes were relative to agiven starting value. (A = 200, c = 10, etc.)We can take the ideas one step further and create a linear function that approxi-mates our given (usually non-linear) function.

Unit 4 - 16

Example: Let us return to the advertising problem, where R = f(A) rep-resents the revenue of a company, given the amount A spent on advertising.Suppose f(200) = 1500, and f ′(200) = 1.8.

State the interpretation of both relationships in words.

Unit 4 - 17

Recall the point/slope form for a linear function:

y = m(x− a) + c

Sketch out the graph of this function, indicating the effect of the parametersm, a and c on the graph.

Unit 4 - 18

Use the point/slope formula, and the information about f(200) and f ′(200), tobuild a local linear approximation for the revenue function R for advertisingbudgets A around 200.

Unit 4 - 19

What revenue would we expect if we reduced advertising to 190 thousand dol-lars?

Unit 4 - 20

We can construct a linear approximation of a function, given a reference pointx = a, using

f(x) ≈ f ′(a)(x− a) + f(a)

This approximation is good assuming that the x values used are “sufficiently” closeto the reference point x = a.The larger (x− a) (or ∆x) is, the worse the approximation will generally be.

Unit 4 - 21

Show that f(x) ≈ f ′(a)(x−a)+f(a) is equivalent to our earlier approximation

f ′(a) ≈∆y

∆x

Unit 4 - 22

Example: Build a local linear approximate formula for the population ofCanada, given it is currently 33 million, and the population is currently in-creasing at a rate 300,000 people per year.

Use your approximation to estimate the Canadian population in 2013.

Unit 4 - 23

Question: Given that the Canadian population is growing exponentially (around1% per year), will your 2013 population estimate above an underestimate or anoverestimate of the real population in that year?

(a) Overestimate

(b) Underestimate

Unit 4 - 24

Support your answer with a sketch of the population curve, and the linearapproximation.

Unit 4 - 25

We can also construct interesting geometric questions using tangent lines.

Example: Find the equations of all the lines through the origin that arealso tangent to the parabola (From H-H Section 3.1 #57.)

y = x2 − 2x + 4

Unit 4 - 26

Sketch the parabola and the lines you found.

Unit 4 - 27

Marginal Rates

A fundamental idea in business and finance is themarginal rate of various functions(most commonly taxes, revenue, and costs).

In Ontario, someone earning a salary of $70,000 will pay $15,500 in incometax. What percentage of their income is dedicated to taxes? (We will call thistheir average tax rate.)

If the same person were earning $71,000 they would pay $15,820 in taxes.What is their new average income tax rate?

Unit 4 - 28

In financial reporting, you would hear that this person has a 32% marginal taxrate. Since their average tax rate is clearly lower than 32%, what does thismean?

Unit 4 - 29

In what situations would your marginal tax rate be more important or relevantthan your average tax rate?

Unit 4 - 30

Marginal Tax RateA person’s marginal tax rate is the amount of tax that would be paid on their next$1 of income. It can be written as a percentage or as an amount.

A question you might be asking yourself at this point is “why are we discussingmarginal tax rates in a calculus course, in a unit on derivatives?”

Unit 4 - 31

From the information given about the $70,000 income scenario, sketch a graphof tax payable vs. income for Ontario residents.

What does the marginal tax rate represent on the graph?

Unit 4 - 32

Let T (x) be the amount of tax paid for an income of x dollars. Express themarginal rate using T (x).

Given that $1 is a very small amount relative in financial terms, how couldwe approximate the marginal rate?

Unit 4 - 33

Solving Nonlinear Equations

Example: Solve the equation x2 + 3x− 4 = 0.

Unit 4 - 34

Example: Solve the equation 10e−x = 7

Unit 4 - 35

Example: Solve the equation 10e−x + x = 7

Unit 4 - 36

Newton’s Method

Perhaps surprisingly to some students, there are many relatively simple equationsthat cannot be solved by hand. We look now at a classical numerical methodthat lets us approximate the solution.Note: It is never better to use numerical methods instead of solving by hand, if aby-hand solution is available.

• Numerical solutions are always approximations, not exact.

• By-hand solutions can often be generalized, while numerical solutions have tobe re-calculated if anything changes.

Unit 4 - 37

Re-arrange the equation 10e−x + x = 7 so the RHS is zero.

Call the LHS f(x), and plot a few points of its graph, between x = 0 andx = 4.

Unit 4 - 38

Where could a solution to f(x) = 0 be, based on the points you plotted?

Bonus: what property of f(x) did you use to find the region of the solution?

Unit 4 - 39

Here are a few more points on the graph.

0 1 2 3 4

−4

−3

−2

−1

1

2

3

4

For maybe not-so-obvious reasons, compute the derivative of f(x) at x = 0.5,a point which is close to a root/solution.

Unit 4 - 40

Use the derivative information to sketch the tangent line at x = 0.5 on thezoomed-in graph below.

0 0.1 0.2 0.3 0.4 0.5

−4

−3

−2

−1

1

2

3

4

Sketch the curve on the same graph (lightly, since we don’t know its exactshape).

Would the root of the tangent line be close to the root of the real (curved)function? Why?

Unit 4 - 41

Newton’s Method

1. Convert an equation like g(x) = h(x) into a function on the left hand side:f(x) = g(x)− h(x) = 0

2. Select a starting value of x, x0, near a root of f(x).

3. Use the formula xn+1 = xn−f(xn)

f ′(xn)to find the root of the tangent line at xn.

4. Repeat Step 3 until the xn+1 estimate is sufficiently close to a root.

Unit 4 - 42

Rationale for Step 3 of Newton’s Method: For an arbitrary function,f(x), and a point x = xn, find where a tangent line to f(x) at xn would reachy = 0.

Unit 4 - 43

Apply Step 3 of Newton’s method twice to improve our estimate of the solutionto our original equation 10e−x + x− 7 = 0.

Unit 4 - 44

Evaluate the quality of the x estimate you found.

Unit 4 - 45

Sketch the values we computed on the axes below.

Unit 4 - 46

It can be shown that, under certain common conditions, and a “sufficiently close”initial estimate of the root, Newton’s method will converge very quickly towards anearby root. It will always give just an estimate, though, not an exact answer; asa result, you always have to trade off the amount of work you are willing to do formore steps/increased accuracy.