PH 6 PDF - Calculus Past Paper
Document Details
Uploaded by LavishChalcedony281
University of Texas at Dallas
Tags
Summary
This document contains handwritten notes on calculus, specifically focusing on calculations involving tangent lines, parameterization, and partial derivatives. The document includes worked examples, equations, and various mathematical concepts, likely part of a calculus syllabus.
Full Transcript
# PH #6 ## 1 **X = 2 intersects** $z = f(x,y) = 4x + 3y^2$ **in curve C** **(a) Calculate** $\frac{df}{dy}$ **at (x,y) = (2,3)** $by (x,y) = 24x + 9y^2$ $2(3)(2) + 9(3)^2$ $= 12 + 81 = 93$ **(b) Find parameterization of tangent line to C where y = 3** We need a point and vector! $C = y^2 (...
# PH #6 ## 1 **X = 2 intersects** $z = f(x,y) = 4x + 3y^2$ **in curve C** **(a) Calculate** $\frac{df}{dy}$ **at (x,y) = (2,3)** $by (x,y) = 24x + 9y^2$ $2(3)(2) + 9(3)^2$ $= 12 + 81 = 93$ **(b) Find parameterization of tangent line to C where y = 3** We need a point and vector! $C = y^2 (2) + 34^3$ $C = 2y^2 + 34^3$ $C = 2(3)^2 + 3(3)$ $C = 18 + 81 = 99$ $z = 2y^2 + 34^3$ $z = 2(3)^2 + 34^3$ $z = 18 + 81 = 99$ $P(t) = (2, 3, 99) + t <0, 1, 93>$ $P(t) = p + t\vec{v}$ $P(t) = (2, t, 2t^2 + 3t^3>$ $P'(t) = <0, 1, 4t + 9t^2> $ (vector at y = 3 is tangent) $P'(3) = <0, 1, 4(3) + 9(3)^2>$ $P'(3) = <0, 1, 81> $ $P(3) = (2, 3, 99)$ - The slope is negative when moving along the **x-axis** and also negative when moving along the **y-axis** so $f_{xy} < 0$. ## 2 $U(x,y,z) = ( x^2 + y^2 + z^2)^{3/2}$ $U = (x^2 + Ý^2 + z^2)^{3/2}$ $4x = - \frac{1}{2} *(x^2 + Ý^2 + z^2)^{3/2} * 2x$ $4x = *(x^2 + Ý^2 + z^2)^{3/2} * \frac{2x}{-1*(x^2 + Ý^2 + z^2)^{3/2}}$ $4x = (x^2 + Ý^2 + z^2)^{3/2} * \frac{-2x}{(x^2 + Ý^2 + z^2)^{3/2}}$ $4x = (x^2 + Ý^2 + z^2)^{3/2-3/2} * (- 2x)$ $4x = (x^2 + Ý^2 + z^2)^{0} * (- 2x)$ $4x = (- 2x) * (x^2 + Ý^2 + z^2)^{0}$ $4x = (- 2x) * 1 $4x = - 2x$ $4xx^2 = 2x^2 * ( x^2 + Ý^2 + z^2)^{3/2}$ $4xx^2 * \frac{ -(x^2 + Ý^2 + z^2)^{3/2}}{(x^2 + Ý^2 + z^2)^{5/2}} = 2x^2 * ( x^2 + Ý^2 + z^2)^{3/2} * \frac{ -(x^2 + Ý^2 + z^2)^{3/2}}{(x^2 + Ý^2 + z^2)^{5/2}}$ $4xx^2 * ( x^2 + Ý^2 + z^2)^{-1} = -2x^2 * ( x^2 + Ý^2 + z^2)^{3/2 - 5/2} $ $Uxx = \frac{4x^2}{(x^2 + Ý^2 + z^2)} - \frac{2x^2}{(x^2 + Ý^2 + z^2)}$ $Uyy = \frac{3y^2(2y)}{(x^2 + Ý^2 + z^2)^{3/2}} - (x^2 + Ý^2 + z^2)^{3/2} * \frac{3y^2(2y)}{(x^2 + Ý^2 + z^2)^{5/2}}$ $Uyy = \frac{6y^3}{(x^2 + Ý^2 + z^2)^{3/2}} - \frac{6y^3}{(x^2 + Ý^2 + z^2)^{3/2}} $ $Uzz = \frac{3z^2(2z)}{(x^2 + Ý^2 + z^2)^{3/2}} - (x^2 + Ý^2 + z^2)^{3/2} * \frac{3z^2(2z)}{(x^2 + Ý^2 + z^2)^{5/2}}$ $Uzz = \frac{6z^3}{(x^2 + Ý^2 + z^2)^{3/2}} - \frac{6z^3}{(x^2 + Ý^2 + z^2)^{3/2}} $ $Uxx + Uyy + Uzz = \frac{4x^2}{(x^2 + Ý^2 + z^2)} - \frac{2x^2}{(x^2 + Ý^2 + z^2)} + \frac{6y^3}{(x^2 + Ý^2 + z^2)^{3/2}} - \frac{6y^3}{(x^2 + Ý^2 + z^2)^{3/2}} + \frac{6z^3}{(x^2 + Ý^2 + z^2)^{3/2}} - \frac{6z^3}{(x^2 + Ý^2 + z^2)^{3/2}} $ $Uxx + Uyy + Uzz = \frac{4x^2}{(x^2 + Ý^2 + z^2)} - \frac{2x^2}{(x^2 + Ý^2 + z^2)} = 0$ $(\frac{x^2}{(x^2 + Ý^2 + z^2)}) * \frac{ 2 }{(x^2 + Ý^2 + z^2)^{3/2}} = 0$ $\frac{2x^2}{(x^2 + Ý^2 + z^2)^{5/2}} =0$ ## 3 **(a) Equation of tangent plane at (3,-4,5)** $f(x,y) = \sqrt{x^2 + y^2} = 2$ - $z = f(a,b) + f_x (a,b) (x-a) + f_y (a,b)(y-b)$ - $f_x = \frac{1}{2} * (x^2 + y^2)^{3/2} * 2x = \frac{x}{(x^2 + y^2)^{1/2}}$ - $f_y = \frac{1}{2} *(x^2 + y^2)^{-1/2} * 2y = \frac{y}{(x^2 + y^2)^{1/2}}$ Plug in (3, -4) for both - $f_x (3, -4) = \frac{3}{5}$ - $f_y (3, -4) = -\frac{4}{5}$ $z = 5 + \frac{3}{5} *(x - 3) + -\frac{4}{5} *(y+4) $ $z = x - \frac{3}{5} + 5 + -\frac{4}{5}y - \frac{16}{5}$ $z = x - \frac{4}{5}y + \frac{14}{5}$ **(b)** $L(x,y) = 5 + \frac{3}{5}(3.1 - 3) - \frac{4}{5}(3.8 + 4) $ $L(x,y) = 5 + \frac{3}{5}(.1) - \frac{4}{5}(.2)$ $L(x,y) = 5 + .06 - .16 $ $L(x,y) = 4.9$ ## 4 $L(x,y) = 6 + 5(3.1 - 3) - 2(3.9-4) $ $L(x,y) = 6 + .5 - 1.2$ $L(x,y) = 6.7$ - Previous given $x$ and $y$ - $L(x,y) = f(x,y) + f_x(x-a) + f_y(y-b)$ - Plug in $(3.1, 3.9)$