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MATH 241 Lecture Notes - Lecture 5: Tangent Space, Partial Derivative

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sapphiretiger55

11 Jul 2017

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University of Maryland

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Course

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Roohollah Ebrahimian

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The equation of the tangent plane of a particular function is. Ok, first we need to compute the partial derivatives of this function. (cid:1878)=(cid:4666)(cid:1876)0,(cid:1877)0(cid:4667)+(cid:3051)(cid:4666)(cid:1876)0,(cid:1877)0(cid:4667)(cid:4666)(cid:1876) (cid:1876)0(cid:4667)+(cid:3052)(cid:4666)(cid:1876)0,(cid:1877)0(cid:4667)(cid:4666)(cid:1877) (cid:1877)0(cid:4667) Where (cid:3051),(cid:3052) are the partial derivatives of that function: find the equation of the tangent plane to (cid:1878)=(cid:884)(cid:3051)ln (cid:4666)(cid:1877)(cid:4667) at (cid:4666)(cid:883),(cid:886)(cid:4667) (cid:3051)=ln(cid:4666)(cid:884)(cid:4667) (cid:884)(cid:3051)ln (cid:4666)(cid:1877)(cid:4667) (cid:3052)=(cid:884)(cid:3051)(cid:1877) (cid:3051)(cid:4666)(cid:883),(cid:886)(cid:4667)=(cid:884)ln(cid:4666)(cid:884)(cid:4667)ln(cid:4666)(cid:886)(cid:4667) (cid:3052)(cid:4666)(cid:883),(cid:886)(cid:4667)=(cid:883)(cid:884) (cid:1878)0=(cid:884) ln (cid:4666)(cid:886)(cid:4667) And now we need to evaluate the partial derivatives at the given point. So the equation of the tangent plane is (cid:1878)=(cid:884) ln(cid:4666)(cid:886)(cid:4667)+(cid:884)ln(cid:4666)(cid:884)(cid:4667)ln(cid:4666)(cid:886)(cid:4667)(cid:4666)(cid:1876) (cid:883)(cid:4667)+(cid:883)(cid:884)(cid:4666)(cid:1877) (cid:886)(cid:4667: find the equation of the tangent plane to (cid:1878)=ysin(cid:4666)(cid:1876)(cid:4667)+ 2(cid:3051)(cid:3052) at (4,6) We find the partial derivatives as usual. (cid:3051)=(cid:1877)(cid:4666)(cid:1876)(cid:4667)+(cid:884)(cid:1877)ln (cid:4666)(cid:1876)(cid:1877)(cid:4667) (cid:3052)=sin(cid:4666)(cid:1876)(cid:4667)+(cid:884)(cid:1876)ln (cid:4666)(cid:1876)(cid:1877)(cid:4667) And now we find the values at the point given. And now we find (cid:1878)0 (cid:3051)(cid:4666)(cid:886),(cid:888)(cid:4667)=(cid:888)+(cid:883)(cid:884)ln (cid:4666)(cid:884)(cid:886)(cid:4667) (cid:3052)=(cid:890)ln (cid:4666)(cid:884)(cid:886)(cid:4667) (cid:1878)0=(cid:883)(cid:883)(cid:884) So the final equation of the tangent plane is (cid:1878)=(cid:883)(cid:883)(cid:884)+(cid:4666)(cid:888)+(cid:883)(cid:884)ln(cid:4666)(cid:884)(cid:886)(cid:4667)(cid:4666)(cid:1876) (cid:886)(cid:4667)+(cid:890)ln (cid:4666)(cid:884)(cid:886)(cid:4667)(cid:4666)(cid:1877) (cid:888)(cid:4667) Since the deviations from the actual point are very small, the linear approximation is a good estimate of the true value.

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Related Questions

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This exercise is intended to give you some insight into why the second derivative test works. A good pan of what I'm asking you to do has been adapted form the Discovery Project on page 821 of our text. In 11.4, we wrote down the tangent plane or linear approximation of a function f(x, y) at a point. To simplify this exercise, we're going to assume that the point is located at the origin, x=0, y=0. Then At a critical point, the two partial derivatives are 0, so the linear approximation just simplifies f(x, y) f(0,0), which is the equation of a horizontal plane. This just doesn't give enough information to determine whether we have a max, min, or saddle. So we look at the quadratic approximation at (0,0), which is defined as follows: At a critical point, this will reduce to It can be proved that if the function f(x, y) has continuous second partials, then it will have the same type of critical point as its quadratic approximation. Let f(x, y) = e-x2-2y2 Find the first partial derivatives Find the first partial derivatives fx(x, y) and fy(x, y): Evaluate fx(0,0) and fy(0, 0). Find the second partial derivatives fxx(x, y), fxy(x, y) and fyy(x, y): Evaluate fxx, fxy and fyy at the critical point. In order to recognize the shape of this graph from its equation, it will be helpful to rewrite the equation in a different form so that: Complete the square for the quadratic approximation and state the values of a, b, c, and d. What docs the graph of the quadratic approximation look like? A: paraboloid opening up. B: paraboloid opening down, C: saddle. Judging from your knowledge of the shape of the graph in exercise 16, does the quadratic approximation have a max, min or saddle? Enter MAX, MIN, or SADDLE. Recall that in the second derivative test, at a critical point (a. b). For f(x, y), state the values of D and fxx. Does the second derivative test tell you that the critical point of f is a local maximum, local minimum, or saddle? After you've completed the lab, we'll take time out to see how the quadratic approximation relates to the second derivative test.

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)