bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§2.6 Implicit

Differentiation



Review §

Any QUESTIONS About• §2.5 → MarginalAnalysis and

Increments

Any QUESTIONS About HomeWork• §2.5 → HW-11

2.5

http://www.google.com/url?sa=i&rct=j&q=delta+x&source=images&cd=&cad=rja&docid=77huDWkZza8kAM&tbnid=urW2yhtXjrCYmM:&ved=0CAUQjRw&url=http://www.haverford.edu/astronomy/songs/lehrer/deriv_physics.htm&ei=5traUcWRFeaziwLH8IHIBw&bvm=bv.48705608,d.cGE&psig=AFQjCNEU8OZi3wpXabydoK25J1hHaTy0og&ust=1373383717111817



§2.6 Learning Goals

Use implicit differentiation to find slopes and Rates of Change

Examine applied problems involving related rates of change

http://www.google.com/url?sa=i&rct=j&q=implicit+differentiation&source=images&cd=&cad=rja&docid=TdjrYwVeOxiZ_M&tbnid=hkykIyLxpKmlNM:&ved=0CAUQjRw&url=http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=ma&chap_sec=02.5&page=theory&ei=J97aUfbxDM3ligKi2IGgDQ&bvm=bv.48705608,d.cGE&psig=AFQjCNFRwEMu8Cce7jOdVsGDY0N8mdLwYA&ust=1373384314906060



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 ''''



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?



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




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



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



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

http://www.google.com/url?sa=i&rct=j&q=implicit+differentiation&source=images&cd=&cad=rja&docid=PNa4OSPjv7YZuM&tbnid=47E4f4keqM_7HM:&ved=0CAUQjRw&url=http://wowmath.org/Calculus/CalculusNotes.html&ei=yunaUYWyAeWliQLup4CADg&bvm=bv.48705608,d.cGE&psig=AFQjCNFRwEMu8Cce7jOdVsGDY0N8mdLwYA&ust=1373384314906060



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




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”.



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




SOLUTION - Comparison Although the answers to parts (a) and

(b) may look different, they should (and DO) agree:• Part (a)

• Part (b)



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?

http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=_piBXPnHNImziM&tbnid=DJgGIxlUrCa-yM:&ved=0CAUQjRw&url=http://www.nanographics.biz/serv01.htm&ei=WznbUfL6ENDwiQKfrID4DA&psig=AFQjCNEDg-V-frEd3oMFF2UCMKZQO33q-A&ust=1373407948517669

http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=o_rv8u8JX2L9YM&tbnid=oVRuzBdgDprXYM:&ved=&url=http://www.webelements.com/compounds/sodium/sodium_chloride.html&ei=jDnbUZeVJKmqiALbxoCwBw&psig=AFQjCNHorNKmUsT_nsK5xEbjSXeMZFIeQg&ust=1373408012663559




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?




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




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




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

http://www.google.com/url?sa=i&rct=j&q=salt+crystal&source=images&cd=&cad=rja&docid=dmV7ZmBmeQsVbM&tbnid=8Y3jZUeP9rei0M:&ved=0CAUQjRw&url=http://stoichiometricequiv.blogspot.com/2012/11/foodchem-carnival-i-am-most-thankful.html&ei=wz7bUaSQFoS7igKXk4GQDQ&bvm=bv.48705608,d.cGE&psig=AFQjCNFlqlSJpFAxAEjUOfWq93vJvFjtfA&ust=1373407896151402



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



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



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



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



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



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



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




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




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




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




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)




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



WhiteBoard Work

Problems From §2.6• P44 → Manufacturing Input-Compensation• P58 → Adiabatic Chemistry• P60 → Melting Ice

http://www.google.com/url?sa=i&rct=j&q=manufacturing+inputs&source=images&cd=&cad=rja&docid=6wiKieCHcynzLM&tbnid=qyF3On4VHgC9DM:&ved=0CAUQjRw&url=http://actionplan.gc.ca/en/initiative/tariff-relief-manufacturing-inputs-and-machinery&ei=lmjcUe39FMXCigKjpYH4CQ&bvm=bv.48705608,d.cGE&psig=AFQjCNEYd23zCVPuCwq0f4qJYOrOcRWQDA&ust=1373485449293082

http://www.google.com/url?sa=i&rct=j&q=melting+ice+cube&source=images&cd=&cad=rja&docid=j-9UdqjxuNHKuM&tbnid=LJwj5LJCZyTbHM:&ved=0CAUQjRw&url=http://wotpast.wordpress.com/2012/07/02/motivate-me-day-two-melting-ice/&ei=ANncUevvHMe9iwL3ioDwCQ&bvm=bv.48705608,d.cGE&psig=AFQjCNHqUwZxMsK-a1hrfgTscWWd5u_qpA&ust=1373514342549894



All Done for Today

IUnderstandImplicitly

http://www.google.com/url?sa=i&rct=j&q=implicit+differentiation&source=images&cd=&cad=rja&docid=-pM1sO_6siWT_M&tbnid=z46iqFM94TfeLM:&ved=0CAUQjRw&url=http://education-portal.com/academy/lesson/how-to-find-derivatives-of-implicit-functions.html&ei=493aUb2PL83ligKi2IGgDQ&bvm=bv.48705608,d.cGE&psig=AFQjCNFRwEMu8Cce7jOdVsGDY0N8mdLwYA&ust=1373384314906060



P2.6-44



P2.6-58



http://www.google.com/url?sa=i&rct=j&q=derivative+power+rule&source=images&cd=&cad=rja&docid=NjbfKj_JZBSzwM&tbnid=jJkxz7X095T-NM:&ved=0CAUQjRw&url=http://mathworld.wolfram.com/Derivative.html&ei=TfPWUe_nA6akiQKZ64DIDw&bvm=bv.48705608,d.cGE&psig=AFQjCNGw-AM0Kc2tARyUGs-zPM7iK-qH3g&ust=1373127702659906

http://www.google.com/url?sa=i&rct=j&q=implicit+differentiation&source=images&cd=&cad=rja&docid=-pM1sO_6siWT_M&tbnid=z46iqFM94TfeLM:&ved=0CAUQjRw&url=http://education-portal.com/academy/lesson/how-to-find-derivatives-of-implicit-functions.html&ei=493aUb2PL83ligKi2IGgDQ&bvm=bv.48705608,d.cGE&psig=AFQjCNFRwEMu8Cce7jOdVsGDY0N8mdLwYA&ust=1373384314906060

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=wSGtOAIrRxIoIM&tbnid=l4Q1f5xFVNRraM:&ved=0CAUQjRw&url=http://www.cantonschools.org/~lforastiere/ap%20calculus?Templates=Printable&ei=2TLcUY-gOa2UjALp4IHgDw&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=nuuUeFKHM5pqjM&tbnid=x7raEhpCGJ8edM:&ved=0CAUQjRw&url=http://designatedderiver.wikispaces.com/Games+and+fun+stuff&ei=IzPcUYPxBaOLiwLjs4GYAQ&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

Documents

pe chabot college mathematics2

derivative of

xmax setgca

axisxmin xmax ymin ymax

equation 2x2 y2

xmin xmax zyh

differentiable fcn of

n s j o h n s o n c