Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Chabot Mathematics §2.6 ImplicitDifferentiation Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. 2.5 Review § • Any QUESTIONS About • §2.5 → MarginalAnalysis and Increments • Any QUESTIONS About HomeWork • §2.5 → HW-11

3. §2.6 Learning Goals • Use implicit differentiation to find slopesand Rates of Change • Examine applied problems involving related rates of change

4. 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.

5. 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 isdy/dx?

6. Comparison: Implicit vs Direct • In thex·y(x) + [y(x)]1/2 = x3 − 23 Problemy(x) could NOT beisolatedalgebraically; we HAD to use ImpilicitDifferentiationtofinddy/dx • Sometimes, however, thereis a choice • Consider the equation 2x2 + y2 = 8, the graph of which is an ellipsein the xy-plane

7. Comparison: Implicit vs Direct • For the Expression 2x2 + y2 = 8 • Compute dy/dx by isolatingy 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

8. % 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 = [xminxmax]; zyh = [0 0]; zxv = [0 0]; zyv = [yminymax]; % % 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([xminxmaxyminymax]),... 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

9. 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 onWhiteBoard

10. Comparison: Implicit vs Direct • SOLUTION (a) • First Isolate y: • Now differentiate with respect to x: • Thus Ans

11. 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”.

12. 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

13. 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)

14. 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?

15. 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?

16. 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):

17. 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

18. 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.

19. 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

20. 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

21. Related Rates • Employ the Chain Rule as • Note that • Thus now have numbers for both dr/dtand dt/dr

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

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

24. Related Rates • Recall • So

25. 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)

26. 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

27. 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 xin Favor of t

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

29. 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)

30. 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)

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

32. All Done for Today IUnderstandImplicitly

41. P2.6-44

43. P2.6-58