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Do Homework- dustin Massengill-Google Chrome Secure https://www.mathxd.com/Student/PlayerHomework.aspx?homeworkld 4424187218ques... FALL 2017 MAT 271 Homework: 3.7, 11.2 Score: 0 of 1 pt 11.2.12 Save 4) 18 of 18 (17 complete) HHW Score: 83.33%, 15 o Question Help Find an equation for the line tangent to the curve at the point defined by the given value of t Also, find the value of at this point x -cos t, y-5+sint, t The equation for the line tangent to the curve at t isy (Simplity your answer Type your answer in slope-intercept form ) Enter your answer in the answer box and then click Check Answer Check Answer Clea emaining 10:20 PM ^9/20/2017

Explanation / Answer

11.2.12)

x=-cost , y=5+sint , t =/2

dx/dt=sint , dy/dt=0+cost

dx/dt=sint , dy/dt=cost

(dy/dx)=(dy/dt)/(dx/dt)

(dy/dx)=(cost)/(sint)

(dy/dx)=(cot t)

at ,t=/2

x=-cos(/2) =0 , y=5+sin(/2)=6

point on curves is (0,6)

dx/dt=sin(/2)=1 , dy/dt=cos(/2)=0

slope of tangent ,(dy/dx)=(dy/dt)/(dx/dt)

(dy/dx)=(0/1)

(dy/dx)=0

now, equation of tangent with slope 0 at point (0,6) is y=6

dx/dt=sint , dy/dx=cott

dx/dt=sint , d/dt(dy/dx)=-csc2t

d2y/dx2=( d/dt(dy/dx))/(dx/dt)

d2y/dx2=-csc2t/sint

d2y/dx2=-csc3t

at t=/2

d2y/dx2=-csc3(/2)

d2y/dx2=-1

===================================================================

x=sect , y=tant , t =-/6

dx/dt=secttant , dy/dt=sec2t

at ,t=-/6

x=sec(-/6) , y=tan(-/6)

x=2/3,y=-1/3

pont on curve is (2/3,-1/3)

dx/dt=sec(-/6)tan(-/6) =(-2/3), dy/dt=sec2(-/6) =(4/3)

slope of tangent ,(dy/dx)=(dy/dt)/(dx/dt)

(dy/dx)=(4/3)/(-2/3)

(dy/dx)=-2

now, equation of tangent with slope -2 at point (2/3,-1/3) is

y-(-1/3) =-2(x-(2/3))

=>y =-2x+(4/3)-(1/3)

=>y =-2x+3

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