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Multivariable Calculus Summary 1 - Partial Derivatives Limits: when dealing with a function of two variables, we see that (x,y) can approach (a,b) along many different paths. In order for a limit to exist, we must get the same value for the limit no matter what path is used in a approaching (a,b). A convenient method to show that a given limit does not exist is to show that we get different results when we approach (a, b) along different paths. Note: limits of continuous functions can be found by substitution provided the point in question is a limit point of the domain of the function. Example: Show that lim x0 y0 x2 y Proof: lim x 0 y 0 x4 y2 x2y x 4 y2 lim x 0 y 0 0 along any line y=mx for each m. x 2 mx x 4 (mx) 2 mx3 lim x 4 (m 2 x 2 ) x 0 y 0 lim x 0 y 0 mx x 2 m2 0 2 What is the limit along y x ? Does the above limit exist.? lim x 0 y 0 x2 y x4 y2 lim x 0 y 0 x2 x2 x 4 ( x) 2 lim x 0 y 0 x4 x4 x4 lim x4 x 0 2 x 4 y 0 1 0 2 Therefore, the limit does not exist. u ( x, t ) e n kt sin nx where n and k are constants, u 2u satisfies the one-dimensional heat equation k 2 t x 2 Problem: Show that the function Proof: u e n kt sin nx n 2 k t u 2u n kt e cos nx n and e n kt sin nx n 2 2 x x 2u 1 n kt 1 u 2 e sin nx n 2 k k k t x u 2u k 2 t x 2 2 2 2 -1- m 2 n 2 kt Problem: Show that the function u ( x, y, t ) e the two-dimensional heat equation for an insulated plane sin mx cos ny satisfies 2u 2u u k 2 2 t y x Problem: show that the equation y=sin(x + at) satisfies the one-dimensional wave equation 2 2 y 2 y a t 2 x 2 Problem: Show that the ideal gas law pV=nRT (n is the number of moles of the gas, R is a constant, p is pressure, V is volume, T is temperature) satisfies the equation p V T 1 V T p -2- Problem: Show that the equation equation u( x, y) e x sin y satisfies the Laplace’s 2u 2u 2 0 2 x y Problem: the electric potential field of a point charge q (coulombs) is defined by ( x, y, z ) q x y z 2 2 2 . Show that satisfy the three-dimensional Laplace equation 2 2 2 0 x 2 y 2 z 2 i Problem: Does the equation u ( x, y) ln x 2 y 2 2u 2u 2 0 2 x y -3- satisfy the Laplace’s equation Tangent plane to a surface and normal line: Method I : if z=f(x,y) and the partial derivatives of z are continuous at ( x 0 , y 0 ), then equation of the plane tangent and the line normal are given respectively by: x x 0 y y 0 z z0 a b 1 f f where a ( x0 , y0 ) and b ( x0 , y0 ) x y (z- z 0 ) = a(x- x 0 ) +b(y- y 0 ) and Proof: f f x If z f ( x, y ), then at ( x0 , y0 ), a gives the instantaneous rate of x 1 change of z with respect to x in a fixed y-direction. f for each unit that x changes. x f Therefore if y 2 y1 0 then x2 x1 1 and z 2 z1 a x u 1, 0, a is tangent to the surface at the point ( x0 , y0 ) . Therefore at the given point z is changing a Similarly, f f y gives the instantaneous rate of If z f ( x, y ), then at ( x0 , y0 ), b y 1 change of z with respect to y in a fixed x-direction. f for each unit that x changes. y f Therefore if x2 x1 0 then y2 y1 1 and z 2 z1 b y v 0, 1, b is tangent to the surface at the point ( x0 , y0 ) . This will mean that u v is normal to the plane tangent to the surface z=f(x, y) Therefore at the given point z is changing b i j k u v 1 0 a ai bj k a, b, 1 is normal to the plane 0 1 b Therefore, the equation of the plane is given by -4- a( x x0 ) b( y y0 1( z z0 ) 0 or ( z z0 ) a( x x0 ) b( y y0 where a f f and b at ( x0 , y0 ) x y Method II : if F(x,y, z)=0) and the partial derivatives of F are continuous at ( x 0 , y 0 , z0 ), then equation of the plane tangent and the line normal are given respectively by: a(x- x 0 ) +b(y- y 0 ) +c (z- z 0 )=0 and where a x x 0 y y 0 z z0 a b c F F F ( x0 , y0 , z0 ) and b ( x0 , y0 , z0 ) and c ( x0 , y0 , z0 ) x y z Tangent line and normal plane to curve If x=f(t), y=g(t) and z=h(t) are parametric equations of a curve in space, then the equation of the normal plane and the tangent line at the point ( x 0 , y 0 , z0 ) are given by a(x- x 0 ) +b(y- y 0 ) +c (z- z 0 )=0 and where a x x 0 y y 0 z z0 a b c x y z (t 0 ) and b (t 0 ) and c (t 0 ) t t t Problems 1. Find the equation of the plane tangent and the line normal to the surface z x 2 y 3xy 2 x 3at (1,2,13) Solution: z 2 2 x 2 xy 3 y 3 x 13 at (1, 2, 13) a 13 z x 2 6 xy 13 at (1, 2, 13) b 13 y The equation of the plane is given by -13(x - 1)+13(y + 2) -1(z+13)=0 or 13x -13y + x -26 =0 answer Plane : 13x 13 y z 26 0 Line : x 1 y 2 z 13 13 13 1 2. Find the equation of the plane tangent and the line normal to the surface F ( x, y, z ) x 2 z 4 xy 2 z 3 14 at (1,2,3) -5- Solution: F 2 x 2 xz 4 y 10 at (1, 2, 3) a 10 F y 8 xy 16 at (1, 2, 3) b 16 F 2 2 z x 3 z 28 at (1, 2, 3) c 28 A vector normal to the plane is given by 10, 16, 28 or 5, 8, 14 The equation of the plane is given by 5(x - 1)-8(y + 2) -14(z-3)=0 or 5x -8y -14z +21 =0 answer Plane : 5 x 8 y 14 z 21 0 Line : x 1 y 2 z 3 5 8 14 3. Find the equation of the line tangent and the plane normal to the curve x t 2 3, y 5t 3, andz t 2 2t 5 at t=0 Solution: dx t 2t 0 at t 0 a 0 and x 3 dy 5 at t 0 b 5 and y 3 dt dz 2t 2 2 at t 0 c 2 and z 5 t A vector tan gent to the curve is given by 0, 5, 2 The equation of the plane normal is given by 0(x - 3)+5(y -3) -2(z-5)=0 or 5y -2z- 5 =0 answer Plane : 5 y 2 z 5 0 Line : y 3 z 5 , x 3 5 2 -6-