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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §2.6 Implicit Differentiation
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 2 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 3 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 4 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.
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 5 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 = x 3 − 23 –Then What is dy/dx?
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 6 Bruce Mayer, PE Chabot College Mathematics Comparison: Implicit vs Direct In the x·y(x) + [y(x)] 1/2 = x 3 − 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 2x 2 + y 2 = 8, the graph of which is an ellipse in the xy-plane
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 7 Bruce Mayer, PE Chabot College Mathematics Comparison: Implicit vs Direct For the Expression 2x 2 + y 2 = 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 8 Bruce Mayer, PE Chabot College Mathematics MATLAB Code 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 9 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 Do on White Board
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 10 Bruce Mayer, PE Chabot College Mathematics Comparison: Implicit vs Direct SOLUTION (a) First Isolate y: Now differentiate with respect to x: Thus Ans
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 11 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(y 2 )/dx think of chain rule and how “the square of y” is really “the square of something with x’s in it”.
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 12 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 13 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)
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 14 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?
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 15 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?
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 16 Bruce Mayer, PE Chabot College Mathematics Example Crystal Growth SOLUTION Because the crystal is a cube, we know that V = s 3 Now differentiate the volume equation with respect to time, using the chain rule (because volume and side length both depend on t):
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 17 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 18 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.
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 19 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 20 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 21 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 22 Bruce Mayer, PE Chabot College Mathematics Related Rates Find dS/dr by Direct Differentiation Calc dr/dV by Implicit Differentiation
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 23 Bruce Mayer, PE Chabot College Mathematics Related Rates Solving for dr/dV When r 0 = 52 in, and dr/dt= 3.7 in/min
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 24 Bruce Mayer, PE Chabot College Mathematics Related Rates Recall So
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 25 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)
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 26 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 27 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 28 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
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 29 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 x 0 = x(0.5yr)
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 30 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)
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 31 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §2.6 P44 → Manufacturing Input-Compensation P58 → Adiabatic Chemistry P60 → Melting Ice
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 32 Bruce Mayer, PE Chabot College Mathematics All Done for Today I Understand Implicitly
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 33 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 34 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 35 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 36 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 37 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 38 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 39 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 40 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 41 Bruce Mayer, PE Chabot College Mathematics P2.6-44
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 42 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 43 Bruce Mayer, PE Chabot College Mathematics P2.6-58
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 44 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 45 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 46 Bruce Mayer, PE Chabot College Mathematics
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BMayer@ChabotCollege.edu MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 47 Bruce Mayer, PE Chabot College Mathematics
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