MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §2.6 Implicit Differentiation

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

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

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.

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?

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

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

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([ ]); % 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',[ ], '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

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

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

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”.

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

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)

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?

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?

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):

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

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.

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

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

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

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

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

MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 24 Bruce Mayer, PE Chabot College Mathematics Related Rates  Recall  So

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)

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

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

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

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)

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)

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

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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MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 41 Bruce Mayer, PE Chabot College Mathematics P2.6-44

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MTH15_Lec-11_sec_2-6_Implicit_Diff_.pptx 43 Bruce Mayer, PE Chabot College Mathematics P2.6-58

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