ha lvh262 Homework 13 2 karakurt 56295 This print out should have 11 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Which of the following equations hold for all differentiable vector functions u t v t and all differentiable scalar functions f t A u t v t u t v t u t v t B so the stated equation is FALSE C The derivative of the dot product of two vector functions follows the product rule for derivatives Thus u t v t u t v t u t v t so the stated equation is FALSE keywords derivative scalar function vector function dot product cross product 002 u t v t u t v t u t v t 1 B and C only 2 none of them 3 B only 4 A and B only 5 A and C only r t a tb t5 c 1 r t a b 5t4 c 2 r t b t4 c 3 r t a b 5t4 c 4 r t a b t4 c 5 r t b 5t4 c correct Explanation The derivative of r t a tb t5 c with respect to t is the sum of the derivatives of each of the terms Now d a 0 dt 6 all of them 7 A only correct 8 C only Explanation A The derivative of the cross product of two vector functions follows the product rule for derivatives So the stated equation is TRUE B The derivative of the product of a scalar function and a vector function follows the product rule for derivatives Thus f t u t f t u t f t u t 10 0 points Find the derivative of f t u t f t u t f t u t C 1 d tb b dt while d 5 t c 5t5 1 c 5t4 c dt Consequently r t b 5t4 c keywords vector function derivative 003 10 0 points ha lvh262 Homework 13 2 karakurt 56295 and Determine the vector Z 1 I r t dt Z 0 1 D 0 E 8 2t 6 1 t2 1 t2 1 t 2 1 6 dt 2 0 1 t h 6 i1 3 1 t 0 h t dt when r t 2 Z Consequently 1 I h 8 ln 2 2 6 i I h 2 ln 2 3 i 2 I h 8 ln 2 6 i 3 I h 2 2 3 ln 2 i 004 10 0 points If r t t t4 t5 find r t 4 I h 8 2 6 ln 2 i 5 I h 2 ln 2 3 i correct 1 r t 0 4t2 5t3 6 I h 2 ln 2 3 i 2 r t 1 4t2 5t3 3 r t 1 4t3 5t4 4 r t 0 12t2 20t3 correct 5 r t 1 12t2 20t3 6 r t 0 12t3 20t4 Explanation For a vector function r t hf t g t h t i the components of the vector Z 1 I r t dt 0 are given by Z 1 f t dt 0 Z 1 g t dt 0 Z Explanation For a vector function 1 h t dt respectively But when D 8 E 2t 6 r t 1 t2 1 t2 1 t 2 we see that Z 1 Z f t dt 0 r t hf t g t h t i 0 1 8 dt 2 0 1 t h i1 8 tan 1 t 2 0 the components of the vector function r t are given by f t g t h t respectively But when r t t t4 t5 we see that while Z 0 1 f t 1 1 2t dt g t dt 2 0 1 t h i1 2 ln 1 t ln 2 Z 0 f t 0 while g t 4t3 g t 12t2 ha lvh262 Homework 13 2 karakurt 56295 and h t 5t4 h t 20t3 Consequently keywords tangent line tangent vector vector function derivative vector form parametric form 006 r t 0 12t2 20t3 3 10 0 points Find an equation in vector form for the tangent line at the point 1 3 3 to the space curve given parametrically by r s h cos s 3e2s sin s 3e 2s i keywords 005 10 0 points 1 tangent line h t 3 3t 3 6ti Find an equation in parametric form for the tangent line to the graph of r s h s3 s6 s5 i at the point 1 1 1 2 tangent line h 1 3 6t 3 5ti correct 3 tangent line h 1 3 3t 3 5ti 4 tangent line h t 3 2t 3 5ti 1 x t 3t2 y t 6t5 z t 5t4 5 tangent line h t 3 2t 3 6ti 2 x t 1 2t y t 1 5t z t 1 4t 6 tangent line h 1 3 6t 3 6ti 3 x t 3t 1 y t 6t 1 z t 5t 1 4 x t 1 3t y t 1 6t z t 1 5t 5 x t 2t 1 y t 5t 1 4t 1 6 x t 1 3t y t 1 6t z t 1 5t correct Explanation The graph of r s has tangent vector r s h 3s2 6s5 5s4 i On the other hand 1 1 1 r 1 Thus at 1 1 1 the equation of the tangent line in vector form is L t r 1 tr 1 h 1 3t 1 6t 1 5t i which in parametric form becomes x t 1 3t y t 1 6t z t 1 5t Explanation Since 1 3 3 r 0 the tangent line has vector form r 0 t r 0 But r s h sin s 6e2s cos s 6e 2s i so at r 0 the tangent vector is r 0 h 0 6 1 6 i h 0 6 5 i Consequently at 1 3 3 the tangent line h 1 2 6t 2 5t i 007 10 0 points Find an equation in parametric form for the tangent line to the graph of r s h cos s 3e2s 3e 2s i at the point 1 3 3 1 x t y 3 6t z 3 6t 2 x t y 3 3t z 3 6t ha lvh262 Homework 13 2 karakurt 56295 3 x 0 y 3 2t z 3 2t 1 4 cos 19 4 x 0 y 3 2t z 3 2t 2 5 cos 17 5 x 1 y 3 6t z 3 6t correct 6 x 1 y 3 3t z 3 2t Explanation Note first that 1 3 3 r 0 On the other hand the tangent vector at an arbitrary point r s is given by r s h sin s 6e2s 6e 2s i so at r 0 the tangent vector is r 0 h 0 6 6 i Thus …