Asked by Sarah
An equation of the line tangent to y=sinx+2cosx at (pi/2, 1) is
A. 2x-y=pi-1
B. 2x+y=pi+1
C. 2x-2y=2-pi
D. 4x+2y=2-pi
I got A
A. 2x-y=pi-1
B. 2x+y=pi+1
C. 2x-2y=2-pi
D. 4x+2y=2-pi
I got A
Answers
Answered by
Steve
y' = cosx - 2sinx
y'(pi/2) = 0 - 2
y-1 = -2(x-pi/2)
or
2x+y = 1+pi
I get B.
http://www.wolframalpha.com/input/?i=plot+y%3Dsinx%2B2cosx%2C+y+%3D+1%2Bpi-2x+for+x+%3D+0+to+3
y'(pi/2) = 0 - 2
y-1 = -2(x-pi/2)
or
2x+y = 1+pi
I get B.
http://www.wolframalpha.com/input/?i=plot+y%3Dsinx%2B2cosx%2C+y+%3D+1%2Bpi-2x+for+x+%3D+0+to+3
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