§2.6 Implicit Differentiation Chabot Mathematics §2.6 Implicit Differentiation Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

2.5 Review § Any QUESTIONS About Any QUESTIONS About HomeWork §2.5 → MarginalAnalysis and Increments Any QUESTIONS About HomeWork §2.5 → HW-11

§2.6 Learning Goals Use implicit differentiation to find slopes and Rates of Change Examine applied problems involving related rates of change

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.

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?

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 Do on Doc Camera

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

Comparison: Implicit vs Direct For the Expression 2x2 + y2 = 8 Compute dy/dx by isolating y in the equation and then differentiating 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

MATLAB Code % Bruce Mayer, PE % MTH-15 • 08Jul13 % XYfcnGraph6x6BlueGreenBkGndTemplate1306.m % % The Limits xmin = -2.5; xmax = 2.5; ymin =-3; ymax =3; % The FUNCTION x = linspace(xmin+0.5,xmax-0.5,500); y1 = sqrt(8-2*x.^2); y2 = -sqrt(8-2*x.^2); % The ZERO Lines zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax]; % the 6x6 Plot whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green axes; 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 MATLAB Code

Comparison: Implicit vs Direct SOLUTION (a) First Isolate y: Now differentiate with respect to x using the Chain Rule Thus Ans

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”.

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

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)

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?

Example  Crystal Growth The most challenging part of this question is correctly identifying variables whose value(s) we need and variables whose value(s) we already 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?

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):

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 From Previous slide:

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.

Related Rates 𝑑𝑧 𝑑𝑡 = 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 = 𝑑𝑧 𝑑𝑡 Related rates consist of using the CHAIN RULE BackWards The Chain Rule: Related Rates 𝑑𝑧 𝑑𝑡 = 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 𝑑𝑧 𝑑𝑥 ∙ 𝑑𝑥 𝑑𝑡 = 𝑑𝑧 𝑑𝑡

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

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

Related Rates Employ the Chain Rule as Note that Thus now have numbers for both dr/dt and dt/dr

Related Rates Find dS/dr by Direct Differentiation Calc dr/dV by Implicit Differentiation

Related Rates Solving for dr/dV When r0 = 52 in, and dr/dt= 3.7 in/min

Related Rates Recall So =1 =1

Example  Revenue vs. Time The demand model for a product as a price fcn → Where D ≡ Demand in k-Units (kU) x ≡ Product Price in $k/Unit The price of the item decreases over time as Where: t ≡ Elapsed Time after Product Release in Years (yr)

Example  Revenue vs. Time Given D(x) & x(t) find the rate at which Revenue changes 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

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

Example  Revenue vs. Time Use the ChainRule to determine dR/dt: Or Now Use Product Rule on SqRt Term

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)

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 $1.16M per year (k-Units/year times $k/Unit) k·k=M

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

I Understand Implicitly All Done for Today I Understand Implicitly http://www.webelements.com/webelements/elements/text/Ag/xtal.html

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