Calculus 1000A/B Lecture 23: Calculus 1000 A -Lecture 23- Section 3.5 Continued- Derivatives of Inverse Trigonometric Functions

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Calculus 1000A/B Lecture 22: Calculus 1000 A -Lecture 21- Section 3.5 Implicit Differentiation

Calculus 1000A/B Lecture 23: Calculus 1000 A -Lecture 23- Section 3.5 Continued- Derivatives of Inverse Trigonometric Functions

Calculus 1000A/B Lecture 24: Calculus 1000 A -Lecture 24- Section 3.6 Derivatives of Logarithmic Functions

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Calculus 1000a lecture 23 - section 3. 5 implicit differentiation continued- derivatives of inverse trigonometric functions. Quiz wednesday; sample quiz and sample midterm exam(s) posted; quiz 2 available for pickup. (cid:3052)(cid:3051)=(cid:883: (cid:4666)(cid:2929)i(cid:2924)(cid:3052)(cid:4667) (cid:3052, cos(cid:1877) (cid:1877) =(cid:883, (cid:1877) = (cid:2869)c(cid:2925)(cid:2929)(cid:3052)= (cid:2869) c(cid:2925)(cid:2929)(cid:4666)(cid:2929)i(cid:2924) (cid:3117)(cid:3051)(cid:4667) sin(cid:4666)sin (cid:2869)(cid:1876)(cid:4667)=(cid:1876) (cid:4666)cos(cid:1876)(cid:4667)(cid:2870)+(cid:4666)sin(cid:1876)(cid:4667)(cid:2870)=(cid:883) cos(cid:1876)= (cid:883) (cid:4666)sin(cid:1876)(cid:4667)(cid:2870) If we look at the graphs of sine and cosine, we will see that, on the interval where sin y is 1:1, cosine is positive, therefore we will take the positive root. (cid:2869) [(cid:2929)i(cid:2924)(cid:4666)(cid:2929)i(cid:2924) (cid:3117)(cid:3051)(cid:4667)](cid:3118)= (cid:2869) (cid:2869) (cid:3051)(cid:3118) (cid:2869: (cid:1877) = (cid:2869)c(cid:2925)(cid:2929)(cid:3052), we can use the same process as before, (cid:1877)=cos (cid:2869)(cid:1876) cos(cid:1877)=(cid:1876, this is if cos y is on the interval (cid:882) (cid:1877) - this is where the function is 1:1 (cid:3051)(cid:4666)cos(cid:1877)(cid:4667)= (cid:3051)(cid:4666)(cid:1876)(cid:4667, (cid:4666)c(cid:2925)(cid:2929)(cid:3052)(cid:4667) (cid:2869) [c(cid:2925)(cid:2929)(cid:4666)c(cid:2925)(cid:2929) (cid:3117)(cid:3051)(cid:4667)](cid:3118)= (cid:2869) (cid:2869) (cid:3051)(cid:3118) (cid:2869: (cid:1877) = (cid:2869)(cid:2929)i(cid:2924)(cid:3052)= , (cid:1877) = (cid:2869)(cid:2929)i(cid:2924)(cid:3052)= (cid:2869) (cid:2929)i(cid:2924)(cid:4666)c(cid:2925)(cid:2929) (cid:3117)(cid:3051)(cid:4667) sin(cid:1877)= (cid:883) (cid:4666)cos(cid:1876)(cid:4667)(cid:2870)

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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

Calculus 1000A/B Lecture 23: Calculus 1000 A -Lecture 23- Section 3.5 Continued- Derivatives of Inverse Trigonometric Functions

Calculus 1000A/B Lecture 22: Calculus 1000 A -Lecture 21- Section 3.5 Implicit Differentiation

Calculus 1000A/B Lecture 23: Calculus 1000 A -Lecture 23- Section 3.5 Continued- Derivatives of Inverse Trigonometric Functions

Calculus 1000A/B Lecture 24: Calculus 1000 A -Lecture 24- Section 3.6 Derivatives of Logarithmic Functions

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