bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege
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BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
§2.6 Implicit
Differentiation
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx2
Bruce Mayer, PE Chabot College Mathematics
Review §
Any QUESTIONS About• §2.5 → MarginalAnalysis and
Increments
Any QUESTIONS About HomeWork• §2.5 → HW-11
2.5
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx3
Bruce Mayer, PE Chabot College Mathematics
§2.6 Learning Goals
Use implicit differentiation to find slopes and Rates of Change
Examine applied problems involving related rates of change
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx4
Bruce Mayer, PE Chabot College Mathematics
ReCall the Chain Rule
If f(u) is a differentiable fcn of u, and u(x) is a differentiable fcn of x, then
That is, the derivative of the composite function is the derivative of the “outside” function times the derivative of the “inside” function.
dx
df
dx
du
du
dfxuufxufxf ''''
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx5
Bruce Mayer, PE Chabot College Mathematics
Implicit Differentiation
Implicit differentiation is the process of computing the derivative of the terms on BOTH sides of an equation.
This method is usually employed to find the derivative of a dependent variable when it is difficult or impossible to isolate the dependent variable itself.• This Typically Occurs for MULTIvariable
expressions; e.g., x·y(x) + [y(x)]1/2 = x3 − 23– Then What is dy/dx?
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx6
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct
In the x·y(x) + [y(x)]1/2 = x3 − 23 Problem y(x) could NOT be isolated algebraically; we HAD to use Impilicit Differentiation to find dy/dx• Sometimes, however, there is a choice
Consider the equation 2x2 + y2 = 8, the graph of which is an ellipse in the xy-plane
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-3
-2
-1
0
1
2
3
x
y
MTH15 • 2x2 + y2 = 8 Ellipse
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx7
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct
For the Expression 2x2 + y2 = 8 a) Compute dy/dx by isolating y in the
equation and then differentiating
b) Compute dy/dx by differentiating each term in the equation with respect to x and then solving for the derivative of y.
Compare the Two Results
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-3
-2
-1
0
1
2
3
x
y
MTH15 • 2x2 + y2 = 8 Ellipse
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx8
Bruce Mayer, PE Chabot College Mathematics
MA
TL
AB
Co
de
% Bruce Mayer, PE% MTH-15 • 08Jul13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%% The Limitsxmin = -2.5; xmax = 2.5; ymin =-3; ymax =3;% The FUNCTIONx = linspace(xmin+0.5,xmax-0.5,500); y1 = sqrt(8-2*x.^2); y2 = -sqrt(8-2*x.^2);% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotwhitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenaxes; set(gca,'FontSize',12);plot(x,y1,'b', x,y2, 'b', 'LineWidth', 4),axis([xmin xmax ymin ymax]),... grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y'),... title(['\fontsize{16}MTH15 • 2x^2 + y^2 = 8 Ellipse',]),... annotation('textbox',[.15 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', ' ','FontSize',7)hold on set(gca,'XTick',[xmin:.5:xmax]); set(gca,'YTick',[ymin:1:ymax])plot([0,0],[ymin,ymax], 'k', [xmin, xmax], [0,0], 'k', 'LineWidth', 2)hold off
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx9
Bruce Mayer, PE Chabot College Mathematics
Example Implicit Differentiation
If y = y(x) Then Find dy/dx from:
y(x) can NOT be algebraically isolated in this Expression (darn!)• Work-Around the Lack of
Isolation using IMPLICIT Differentiation
233 xyxxyx
23
3
xy
xxy
x
dxd
Do onWhiteBoard
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx10
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct SOLUTION (a) First Isolate y:
Now differentiate with respect to x:
Thus Ans
xx 4282
1 2/12
2/1228
2
x
x
228
2
x
x
dx
dy
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx11
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct
SOLUTION (b) This last step is where the challenge (and
value) of implicit differentiation arises. Each term is differentiated with x as its input, so we carefully consider that y is itself an expression that depends on x• Thus, when we compute d(y2)/dx think of
chain rule and how “the square of y” is really “the square of something with x’s in it”.
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx12
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct Using the implicit
differentiation strategy, first differentiate each term in the equation:
Then
Now solve for the dy/dx term
Thus Ans
y
x
dx
dy
2
4 04 2
dx
dyy
dy
dx
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx13
Bruce Mayer, PE Chabot College Mathematics
Comparison: Implicit vs Direct
SOLUTION - Comparison Although the answers to parts (a) and
(b) may look different, they should (and DO) agree:• Part (a)
• Part (b)
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx14
Bruce Mayer, PE Chabot College Mathematics
Example Crystal Growth
A sodium chloride crystal (c.f. ENGR45) grows in the shape of a cube, with its side lengths increasing by about 0.3 mm per hour.
At what rate does the volume of the rock salt crystal grow with respect to time when the cube is 3 mm on a side?
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx15
Bruce Mayer, PE Chabot College Mathematics
Example Crystal Growth
The most challenging part of this question is correctly identifying variables whose value we need and variables whose value we know.
First, carefully examine the question
At what rate does the volume of the rock salt crystal grow with respect to time when the cube is 3 mm on a side?
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx16
Bruce Mayer, PE Chabot College Mathematics
Example Crystal Growth
SOLUTION Because the crystal is a cube, we know
that V = s3
Now differentiate the volume equation with respect to time, using the chain rule (because volume and side length both depend on t):
3sdt
dV
dt
d
tsdt
ds
dt
dV 23
dt
dss
ds
dV
dt
d 3
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx17
Bruce Mayer, PE Chabot College Mathematics
Example Crystal Growth
Need to Evaluate dV/dt when s = 3 Recall that the side length is growing at
an instantaneous rate of 0.3 mm per hour; that is:
Then since
3.0dt
ds
dt
dssts
dt
ds
dt
dV 22 33
Hr
mm 1.83.033
32
3.0
3
dtds
sdt
dV
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx18
Bruce Mayer, PE Chabot College Mathematics
Example Crystal Growth
State: When the sides are 3mm long, the sodium Choloride crystal is growing at a rate of 8.1 cubic millimeters per hour.
Hr
mm 1.8
3
3.0
3
dtds
sdt
dV
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx19
Bruce Mayer, PE Chabot College Mathematics
Related Rates
In many situations two, or more, rates (derivatives), are related in Some Way.
Example Consider a Sphere Expanding in TIME with radius, r(t), Surface area, S(t), and Volume, V(t), then
But r, S, and V are related by Geometry
twdt
dVtv
dt
dStu
dt
dr
32
3
44 rVrS
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx20
Bruce Mayer, PE Chabot College Mathematics
Related Rates
Knowing u(t), v(t), and w(t) should allow calculation of quantities such as:
Consider a quick Example.• A 52 inch radius sphere expands at a rate
of 3.7 inch/minute. Find dS/dV for these conditions
• Recognize
dS
dV
dr
dS
dr
dV
min
in 7.3in 52
0
0 rdt
drr
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx21
Bruce Mayer, PE Chabot College Mathematics
Related Rates
Employ the Chain Rule as
Note that
Thus now have numbers for both dr/dt and dt/dr
dV
dr
dr
dt
dt
dr
dr
dS
dV
dt
dt
dS
dV
dS
in
min 2703.0
min 1in 7.3
11
min
in 7.3
drdtdr
dt
dt
dr
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx22
Bruce Mayer, PE Chabot College Mathematics
Related Rates
Find dS/dr by Direct Differentiation
Calc dr/dV by Implicit Differentiation
dr
dSrrrr
dr
dS
dr
d in 8
in
in 8244
22
VrV
dV
dVrV 33
3
4
3
4
33
3
41
3
4r
dV
dVr
dV
dV
dV
d
dV
drr
dV
drr
dr
dr
dV
d 233 33
41
3
41
3
41
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx23
Bruce Mayer, PE Chabot College Mathematics
Related Rates
Solving for dr/dV
When r0 = 52 in, and dr/dt= 3.7 in/min
dV
dr
rdV
drr
22
4
141
in 92.997.388 0
0
rdr
dS
r
23
220 in
1 10813.5
7.34
1
4
1
0
rdV
dr
r
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx24
Bruce Mayer, PE Chabot College Mathematics
Related Rates
Recall
So
dV
dr
dr
dt
dt
dr
dr
dS
dV
dr
dr
dt
dt
dr
dr
dS
dV
dS
23
in
1 10813.5
in
min
7.3
1
min
in
1
7.3
1
in 99.92
0
rdV
dS
in
1 5405.0
in
in 10813.599.92
3
23
0
rdV
dS
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx25
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
The demand model for a product as a function →• Where
–D ≡ Demand in k-Units (kU)– x ≡ Product Price in $k/Unit
The price of the item decreases over time as• Where: t ≡ Time after Product Release in
Years (yr)
124 xxD
26 ttx
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx26
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
Given D(x) & x(t) at what rate is Revenue changing with respect to time six months after the item’s release?
SOLUTION Formalizing the goal with mathematics,
we want to know the rate, dR/dt , six months after release. • Because time is measured in years,
set t = 0.5 years
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx27
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
ReCall Revenue Definition
[Revenue] = [Demand]·[Quantity] Mathematically in this case
The Above states R as fcn of x, but we need dR/dt• Can Use Related-Rates to eliminate x in Favor of t
xDxxR 124 xxx
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx28
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
Use the ChainRule to determine dR/dt:
Or
Now Use Product Rule on SqRt Term
dt
dx
dx
dR
dt
dR
26124 t
dt
dxxx
dx
d
txxdx
d
dt
dR2124
12 xxdx
d 1212122
1 2/1
xxx
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx29
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
Continuing the ReDuction
We need to evaluate the revenue derivative at t = 0.5 yrs, but there’s a catch: We know the value of t, but the value of x is not explicitly known. • Use the Price Fcn to calculate x0 = x(0.5yr)
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx30
Bruce Mayer, PE Chabot College Mathematics
Example Revenue vs. Time
Recall: Then: Can Now Calc dR/dt at the 6mon mark
• State: After 6 months, revenue is increasing at a rate of about $1.162M per year (k-Units/year times $k/Unit)
26 ttx
200 5.065 xtxx 75.5
5.02175.52175.5275.54 2/1
75.5,5.0
xtdt
dR
yr$M 162.1
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx31
Bruce Mayer, PE Chabot College Mathematics
WhiteBoard Work
Problems From §2.6• P44 → Manufacturing Input-Compensation• P58 → Adiabatic Chemistry• P60 → Melting Ice
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx32
Bruce Mayer, PE Chabot College Mathematics
All Done for Today
IUnderstandImplicitly
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx33
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx34
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx35
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx36
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx37
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx38
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx39
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx40
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx41
Bruce Mayer, PE Chabot College Mathematics
P2.6-44
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx42
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx43
Bruce Mayer, PE Chabot College Mathematics
P2.6-58
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx44
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx45
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx46
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx47
Bruce Mayer, PE Chabot College Mathematics
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