exponen’als,,logarithms,and,all, thatkeshet/m102/2015/lect9.1.pdf · similarexample:,...
TRANSCRIPT
Exponen'als, logarithms and all that
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.
Coming up (this week)
UBC Math 102
Group version of Quiz 3
distributed by email
Group version of Quiz 3 due
in class
(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)
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
Anatomy of the exponen'al
UBC Math 102
Prac'ce midterm ques'on
Sketch the graph of the func'on
UBC Math 102
(2) Prac'ce midterm ques'on
My graph should look like:
UBC Math 102
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
Steps in graphing
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)
Exponen'als and logarithms (review) Proper&es of e^x Proper&es of natural logs
Exponen'als and logarithms are inverse func'ons
UBC Math 102
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
Desmos demo:
UBC Math 102
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.
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
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
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
Transformed data
On a log plot, an exponen'al func'on looks like a straight line!
slope = k x
UBC Math 102
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
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!
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
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
Transformed data
On a log-‐log plot, a power func'on looks like a straight line!
y = a xb slope = b
UBC Math 102
Example: Applica'on to allometry: “mice to elephant curve”
Metabolic rate vs animal size.. x y
Data:
UBC Math 102 Examples 10.14-‐.15
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
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
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
Finding a and b:
• A=8.2 = so • Slope = b = -‐ 0.26 Hence: Data fits rela'onship
UBC Math 102
Slope = b = -‐0.26
Exponen'al 'me behaviour, half life and doubling 'mes
UBC Math 102
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
(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
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
(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
Solu'on:
UBC Math 102
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
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
Solu'on:
Note that we get the same equa'on as we did for half-‐life, so same answer! doubling 'me = ln(2)/k
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
Applica'ons:
Human popula'on is increasing (“exponen'al growth”)
UBC Math 102
Mo'va'on for the group-‐quiz this week
Credit:Aspire Food Group
Differen'al equa'ons for exponen'al growth and decay
(preview of next 'me)
UBC Math 102
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
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
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
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)
Answers
1 C 2 A 3 B 4 D 5 A 6 D 7 E 8 A
UBC Math 102
A midterm problem “puzzler”
Compute the deriva've of y=f(x) = xx
UBC Math 102
Solu'ons to exam-‐type problems from previous lecture
UBC Math 102
Prac'ce: Exam ques'on
UBC Math 102
Solu'on
UBC Math 102
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
Solu'on
UBC Math 102
Prac'ce midterm ques'on
Sketch the graph of the func'on
UBC Math 102
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
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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
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
Prac'ce Exam ques'on from last 'me:
UBC Math 102
Solu'on:
UBC Math 102
Solu'on, cont’d
UBC Math 102
Prac'ce Exam ques'on for Implicit differen'a'on from last 'me
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Solu'on:
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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.
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)]