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Exponen'als, logarithms and all that
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Evalua'on -‐ informal
• Please pick up and fill in an (informal) midterm evalua'on form (annonymous)
• This will be used for feedback to improve the way M102 is taught.
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Coming up (this week)
UBC Math 102
Group version of Quiz 3
distributed by email
Group version of Quiz 3 due
in class
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(1) The deriva've of ex is ex
What is the deriva've of y=ekx
(A) ekx
(B) k e(k-‐1)x (C) kekx (D) kekx-‐1 (E) kek(x-‐1)
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The deriva've of ex is ex
By the chain rule, the deriva've of y=ekx
is dy/dx = k ekx
Steps shown hereà
UBC Math 102
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Anatomy of the exponen'al
UBC Math 102
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Prac'ce midterm ques'on
Sketch the graph of the func'on
UBC Math 102
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(2) Prac'ce midterm ques'on
My graph should look like:
UBC Math 102
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If powers and exponen'als “fight” for domina'on .. then
• Exponen'als always win!!
• As x à ∞ x2 à ∞ power term INCREASING
• As x à ∞ e-‐x à0 exponen'al DECREASING
• the product x2 e-‐x à0 DECREASES as x à ∞
UBC Math 102
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Steps in graphing
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The graph looks like:
You can find the cri'cal points, and the value at x=0, small x, and large x. (See details at end of slides)
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Exponen'als and logarithms (review) Proper&es of e^x Proper&es of natural logs
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Exponen'als and logarithms are inverse func'ons
UBC Math 102
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Symmetry proper'es If g(x) is an inverse func'on of f(x) then their graphs are symmetric about line y=x
UBC Math 102
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Desmos demo:
UBC Math 102
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Desmos demo: “the chops'cks”
UBC Math 102
Challenge: see if you can create this demo which shows the tangent lines to ln(x) and exp(x) at corresponding points, demonstra'ng that slopes of these inverse func'ons are reciprocals.
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Applica'on of logarithms to data-‐fi^ng
• Log transforms can help to visualize data that varies on a wide scale
• Log transforms can help “fit data” by conver'ng exponen'al or power rela'onships to linear rela'onships.
UBC Math 102
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Log transforms: exponen'alàlinear
A rela'onship of the form y = C ekx can be expressed as a linear rela'onship between ln(y) and x: y = C ekx
ln(y) = ln(C ekx ) = ln(C) + ln( ekx ) ln(y) = ln(C) + k x
UBC Math 102
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3. Log transforms: exponen'alàlinear
On a plot of ln(y) vs x ln(y) = ln(C) + k x (A) The slope is C and the intercept is k (B) The slope is k and the intercept is ln(C) (C) The slope is k and the intercept is C (D) The slope is –k and the intercept is ln(C) (E) Not sure.
UBC Math 102
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Transformed data
On a log plot, an exponen'al func'on looks like a straight line!
slope = k x
UBC Math 102
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Example: Assignment8: Problem 6
Determine whether the func'on whose values are given in the table below could be linear or exponen'al. Use a spreadsheet: (1) copy and paste the data. Compute ln(y).
UBC Math 102
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Cont’d: plot log(y) vs t and add trend-‐line
UBC Math 102
y = 0.7782x + 0.301
0
0.5
1
1.5
2
2.5
3
3.5
4
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
Log(y)
t
log(y)
log(y)
Linear (log(y)) Careful! This is actually Log(y) vs x!
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Log transforms: power fnàlinear
A rela'onship of the form y = a xb can be expressed as a linear rela'onship between ln(y) and ln(x): y = a xb
ln(y) = ln(a xb ) = ln(a) + ln( xb ) ln(y) = ln(a) + b ln(x)
UBC Math 102
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4.Log transforms: power fnàlinear
On the graph: ln(y) = ln(a) + b ln(x) (A) The slope is ln(x) and the intercept is b (B) The slope is ln(a) and the intercept is b (C) The slope is b and the intercept is ln(y) (D) The slope is b and the intercept is ln(a) (E) Not sure
UBC Math 102
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Transformed data
On a log-‐log plot, a power func'on looks like a straight line!
y = a xb slope = b
UBC Math 102
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Example: Applica'on to allometry: “mice to elephant curve”
Metabolic rate vs animal size.. x y
Data:
UBC Math 102 Examples 10.14-‐.15
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Does data fit a power-‐law?
• Plot ln(y) versus ln(x)
(1) This helps to visualize the data (2) Helps to find values for the constants a and b in
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Fit a straight line to the log-‐log plot
Determine slope and intercept. Use these to compute a,b
UBC Math 102
Slope = b = -‐0.26
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Values of a and b
From this line we find that (A) a=ln(8.2), b=-‐0.26 (B) a=-‐0.26, b=8.2 (C) a=8.2, b=-‐0.26 (D) a=e8.2, b=-‐0.26 (E) Not sure
UBC Math 102
Intercept=8.2
Slope = -‐0.26
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Finding a and b:
• A=8.2 = so • Slope = b = -‐ 0.26 Hence: Data fits rela'onship
UBC Math 102
Slope = b = -‐0.26
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Exponen'al 'me behaviour, half life and doubling 'mes
UBC Math 102
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Exponen'ally decaying process
Consider the exponen'al func'on of 'me:
y = f(t) = C e -‐kt
(with k>0). What is the value of y at 'me t=0? (we will denote that value by y(0)=y0.)
UBC Math 102
Note minus sign in exponent
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(5) Exponen'ally decaying process
For the exponen'al func'on y = f(t) = C e -‐kt
(A) y(0) = y0 = C (B) y(0) = y0 = 0 (C) y(0) = y0 = 1 (D) y(0) = y0 = -‐C k (E) y(0) = y0 = C e -‐k UBC Math 102
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Determine the “half-‐life”
Defini'on: the half-‐life is the 'me at which ½ of the ini'al amount remains, i.e. y(t) = (1/2) y0
UBC Math 102
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(6) The “half-‐life”
Defini'on: the half-‐life is the 'me at which ½ of the ini'al amount remains, i.e. y(t) = (1/2) y0
The half life is (A) C/2 (B) k/2 (C) 2/k
(D) ln(2)/k (E) ln(C)/2k
UBC Math 102
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Solu'on:
UBC Math 102
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Similar Example:
The amount of cesium-‐137, remaining t years aoer the Chernobyl nuclear plant exploded can be modeled by the equa'on
where y0 is the ini'al level. What is the half-‐life? How long would it take for the level of cesium-‐137 to be reduced to 3% of its ini'al level? 0.03 y0 ß solve for t
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Exponen'ally increasing process
Consider the exponen'al func'on of 'me:
y = f(t) = C e kt
(with k>0). The 'me at which y doubles is called the doubling &me. Exercise: show that the doubling 'me is ln(2)/k UBC Math 102
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Solu'on:
Note that we get the same equa'on as we did for half-‐life, so same answer! doubling 'me = ln(2)/k
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Units
• (1) An exponent cannot have units. (Thus, for P(t)=Cert, we conclude that r has units of 1/year).
• (2) e (and therefore ex) does not have dimensions. (Thus, for P(t)=Cert, we can conclude that C has the same units as P(t)).
UBC Math 102
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Applica'ons:
Human popula'on is increasing (“exponen'al growth”)
UBC Math 102
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Mo'va'on for the group-‐quiz this week
Credit:Aspire Food Group
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Differen'al equa'ons for exponen'al growth and decay
(preview of next 'me)
UBC Math 102
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Differen'al equa'on
• The func'on sa'sfies the equa'on:
A differen'al equa'on is an equa'on linking a func'on and its deriva'ves.
UBC Math 102
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Solu'on to a differen'al equa'on
• We say that y=ex is a solu&on to the differen'al equa'on:
dy/dx = y
UBC Math 102
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Solu'on to a differen'al equa'on
• We want to use these facts in 'me-‐dependent systems. Hence the independent variable will usually be t for 'me rather than x.
UBC Math 102
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Solu'on to a differen'al equa'on
• We say that y=ekt is a solu'on to the differen'al equa'on:
dy/dt = ky
UBC Math 102
No'ce: independent variable t ('me)
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Answers
1 C 2 A 3 B 4 D 5 A 6 D 7 E 8 A
UBC Math 102
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A midterm problem “puzzler”
Compute the deriva've of y=f(x) = xx
UBC Math 102
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Solu'ons to exam-‐type problems from previous lecture
UBC Math 102
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Prac'ce: Exam ques'on
UBC Math 102
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Solu'on
UBC Math 102
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Prac'ce: conceptual midterm ques'on Here is the graph of y = C ekt for some constants C, k, and a
tangent line. Use data from the graph to determine C and k.
UBC Math 102
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Solu'on
UBC Math 102
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Prac'ce midterm ques'on
Sketch the graph of the func'on
UBC Math 102
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Solu'on
• At x=0 y=0 • Cri'cal points? y’(x)= dy/dx = 2x e-‐x -‐ x2 e-‐x =0 (2x -‐ x2 )e-‐x =0 • Solve for x: (2x -‐ x2 ) =0 x(2-‐x)=0 so x=0, 2 are cri'cal pts
UBC Math 102
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Solu'on, contd
• Second deriva've y’’(x)= [(2x -‐ x2 )e-‐x ] = (2-‐2x) e-‐x -‐(2x -‐ x2 )e-‐x
= (2-‐4x+x2)e-‐x
At x=0 y’’(0)=2 e0=2>0 concave up (local max) At x=2 y’’(2)=(2-‐8+4)=-‐2 <0 concave down (local min)
UBC Math 102
Product rule
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Solu'on, contd
• Inflec'on points (Ips) y’’(x) = (2-‐4x+x2)e-‐x
y’’(x) changes sign when (2-‐4x+x2)=0 Then, using the quadra'c formula, we find Ips at x=
UBC Math 102
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Prac'ce Exam ques'on from last 'me:
UBC Math 102
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Solu'on:
UBC Math 102
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Solu'on, cont’d
UBC Math 102
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Prac'ce Exam ques'on for Implicit differen'a'on from last 'me
UBC Math 102
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Solu'on:
UBC Math 102
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Solu'on, contd:
UBC Math 102
For part (b) you want to determine the second deriva've to deduce concavity at the given point.
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Prac'ce Exam Ques'on (mul'ple choice)
UBC Math 102
y = log2(x) so dy/dx = 1/[ln(2) x] So at x=1 the slope (deriva've) is 1/[ln(2)]