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MATA36H3 Lecture 2: integration technique

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6 Feb 2019

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MATA36H3 Lecture 1: review of 30

MATA36H3 Lecture 2: integration technique

MATA36H3 Lecture Notes - Lecture 3: Inverse Trigonometric Functions, Fetus

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See a portion of fix ) to be equal to. [ u aux ) ] some quantity involving x. Replace dx for in wad dx the integral dx. After this function do step . not it"s the best include. X that the original like u - ex. If there are still quantities involving x solve x in terms of u. 7 . the new integral , with u being the variable x back into a . get an answer in terms of. 5 no more x f e dx. Given fed , try to set the most ay troublesome mm piece to be u . if ie is stuck inside. U" ox) wcx ) will be the on outside directly multiple to dy will cancel in the process. When the bounds of integral is given c definite integral ) , one can compute new bounds for u. 7 steps , plug in the bounds at the ends.

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1. (45 marks) Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answer. Unless otherwise stated, it is not necessary to simplify your answers in this question. a) Find the equation of the tangent plane to the graph of the function z = f(x,y) = sell at (0,1). Answer b) Evaluate (T, 0) if f(x,y) = sin(ry). Answer c) Graph of a function z = f(x,y) is given by: Which of the following level curves correspondes to this function. Answer d) If f(x,y) = e(3x – 2y), find lim - Math)-f(x,y) e) Find % if f(x, y) = x2 - y3 and 2 = r sin(t), y = r cos(t). Answer f) Evaluate the directional derivative of f(x, y) = !at (1, -1) in the direction of u =(2, 1). Answer 8) Does Simono converge? h) Sketch the direction field of the differential equation = 1-2y. i) Consider the differential equation ay = 15 - 31, y(0) = 0, and use Euler's method with step size 0.1 to estimate y(0.1). Answer j) Knowing that (-3,-3) is a critical point of f(x,y) = 2:3 + y2 – 4.cy -3.0, use the second derivative test to find out whether it is a local minimum, local maximum, or a saddle point. Answer k) Solve the initial value problem, dy dt 642 2y + cos(u) y(1) = . Answer 1) Is ū=(2,2, -1) perpendicular to v = (5,-4, 2)? Answer m) Find the area between the graph of y = 1 + sin(x) and y=1-sin(x) from 0 to 7. Answer n) Determine the lentgh of the curve y = 8/2,0 SIS 3. Evaluate and simplify your answer. Answer o) Express the volume of the solid obtained by rotating the bounded region that lies between curves y = (x-2)2 and y=1,150 $4 about the vertical line 2 = 5. Do not evaluate this integral. Answer

Show transcribed image text 1. 0/6 points LarCalc11 13.1.010. [3864 VIA Evaluate the function at the given values of the independent variables. Simplify the results. fx, y) -4 -x2 - 4y2 (a) ro, 0) for (c) f2, 3) (e) x, o Cal 2. 0/2 points LarCalc11 13.1.019.MI. [386 Evaluate the function at the given values of the independent variables. Simplify the results. x, y)- 6x y2 (a) fix + ax, y)-f(x, y) Ay
3. 1 points LarCalc11 13.2 Find the indicated limit by using the limits limfx, y) 2 and im g(x, y)--4 lim 4. O/1 points LarCalci1 13 2 Find the indicated limit by using the limits lim x, y) 3 and lim (x, y) g(x, y)--5 (a, b) lim x, Y) (x, y) (a, b) g(x, y) 5. y1 points LarCalc11 13.3 Explain whether or not the Quotient Rule should be used to find the partial derivative. Do not differentiate. ay x2 51 O Yes, the function is a quotient. O No, y only occurs in the numerator. 6. 0/2 points LarCalc11 13.3 Find both first partial derivatives oz 7. 0/2 points LarCalc11 13.3 Find both first partial derivatives. hx, y) ex+y) h(x, y)
8. /2 points LarCalc11 13.3.031 Find both first partial derivatives. az dx az ay 9. /1 points LarCalc11 13.4.003. Find the total differential. z 3xSy10 dz 10. 0/2 points LarCalc11 13.5.012 Consider the following. (a) Find dw/dt using the appropriate Chain Rule. dw dt (b) Find dw/dt by converting w to a function of t before differentiating. dw dt 11. 0/1 points LarCalc11 13.6.008. Find the directional derivative of the function at P in the direction of v. rx, y) = x3-y3, Rio, 9), v-V2 (1+j)
12. 0/2 points LarCalc11 13.7.017 (a) Find an equation of the tangent plane to the surface at the given point. x+y+z=9, (2, 5, 2) (b) Find a set of symmetric equations for the normal line to the surface at the given point. ox -2y -5-z-2 13. /3 points LarCalc11 13.8.025 Use a computer algebra system to graph the surface and locate any relative extrema and saddle points. (If an an not exist, enter DNE.) -8x x2 y21 relative minimum (x, y, z) - relative maximum (x, y, z) saddle point 14. 0/1 points LarCalci1 13.10.027 Use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. Surface Point Plane: x yz 1 (7,1,1)

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Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?