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# MATH 121 Exam I Study Guide MATH 121

UVM

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This page Study Guide was uploaded by Henry on Tuesday February 23, 2016. The Study Guide belongs to MATH 121 at University of Vermont taught by Anthony Julianelle in Spring 2016. Since its upload, it has received 50 views. For similar materials see Calculus III in Mathematics (M) at University of Vermont.

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Date Created: 02/23/16

MATH 121 Exam 1 Study Guide Henry Mitchell February 23 2016 1 Finding the angle between two vectors 11 Using the dot product Remember 13 j 1539 cos 6 where 6 is the angle between the two vectors 1539 and f This means that 13 q 17lt l 6 arccos plllt l 12 Using the cross product cos6 Remember x sin6 where 6 is the angle between the two vectors 1539 and f This means that up an 2 sin Plfil 6 arcsin x 17 l 2 Volume of a parallelepiped Remember the volume of a parallelepiped a shape with 6 faces all of them parallelograms is given by the scalar triple product of the vectors which make up three of its edges Remember the scalar triple product of the vectors 15 j and F is the dot product of one of the vectors with the cross product of the other two In other words Remember that as shown above the order in which you perform the operation doesn t matter Those are all equivalent statements 3 Equation of a plane The equation for a plane can be written as am by 02 0 where the vector 0 b c is perpendicular to the plane in question 4 Normal and binormal vectors For a curve de ned by 75 2 lt I3t zt 39r t The unit tangent vector is T T t T tgt39 The unit binormal vector is B t T t X N t won t The unit normal vector is N 5 Parameterization by arc length The goal of parameterization by arc length is to parameterize a function in such a way that puts the particle at a constant speed The way to do this is to nd a function for t in terms of 8 then put the main function in terms of that so we get 7 t8 For the function 7quot the length of the arc of the curve from 0 to t is st0 r udu So we take the inverse of that function dt W and put r in terms of it Honestly this is best explained with an example We re looking for a parameterization in terms of arc length of the following 75 2 cos tsin tt 1 r t sin tcos t 1 2 Hr t sin2t cos2t 12 3 8t zOtllr uduOt du u ut t 4 u0 8t t gt 8 71158 ltcos sin 2 6 1 is our given function 2 is the derivative of that function 3 is the norm of the derivative of the function 4 is nding the arc length as a function of t 5 is nding the inverse of that function nding 75 as a function of arc length 6 is plugging 5 into 6 Curvature Curvature can be de ned in many ways The de nition Which is not particularly useful is dT K d8 However more usefully we also have Ilsa gtlt r tgt Hut H3 and in two dimensions one variable Hm 1 fx232 7 Parametric equations of lines and line segments 71 Lines A line through Aa1 a2 a3 and B b17 b2 b3 has parametric equations t 01 01 1 t W a2 a2 b2 75 7375 a3 a3 b3 t 0139 t 1 01 1 t W 192 a2 b2 75 7375 193 a3 b3 75 72 Line segments A line segment from a point de ned by the vector F0 to a point de ned by the vector F1 has equation 7 05 2 1 t 770 15771 Where 0 g t g 1 8 Proving limits do not exist Remember we are not proving that limits of functions of multiple variables do exist We only have to prove that there exists at least one path along which the limit from one side is not equal to the limit from the other side7 or two paths for which the limit is different Some examples of ways to do this are along the x axis lim w0 gt00 Along the y axis lim 0y gt00 Along y 2 cc lim ww gt00 9 Directional derivatives The derivative of a function f cc7 y in the direction of unit vector 11 a b is a 8 012130373 2 8 3 8 519 10 Chain rule The chain rule in multiple variables works exactly the same as implicit differentiation in one variable There are two cases in which it can be implemented 101 Case 1 Let 2 fcc7 y be differentiable in cc and 3 cc 2 gt7 and y ht This means that z is a differentiable function of t and that dz 8fdcc afdy 013g 3 dt 8 ccdt 8 ydt dt An easier thing to do is simply substitute 9 and h in for cc and 3 giving us 2 as a function of t 102 Case 2 Let z fx 3 cc 2 gst and y hst The derivative of z With respect to t is ampQ at 056 at 8y at 11 Gradient Remember the gradient of a surface at a point is the steepest slope at that area It points in the direction of greatest rate of change and its magnitude is that rate of change a a a f f fk y z fy7 y z fz5v7 y

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