Question

Find the equations of the liens that are tangent and normal to the graph of y = sinx +3 at x= pi

Answer 1

* lines

Answer 2

find dy/dx and set equal to 0
then set dx/dy =0

Answer 3

u have to differentiate it ,then y'=cosx, so when x =pi, y'=-1, this is the tangent slope, so the slope for normal line should be 1, and then plug them in to (pi,3), u can get the equation

Answer 4

the book says \[y = -x + \pi + 3\] i got the y = cosx as my derivative...

Answer 5

cos(π) is the slope of your line.

Answer 6

yea.... but it doesn't say that in the book... idk what i did wrong

Answer 7

Like @mymathcourses said, dy/dx = cos(x). That's the slope of your line.
The line passes through x=π. Putting that into the equation y=sin(x)+3, sin(π)+3 = 3, so the point of intersection is (π,3).
Use
\[y-y_1=m(x-x_1)\]
to solve : m = dy/dx.

Answer 8

yea... but idk how it turns out to the equation in the book

Answer 9

y-3 = cosx (x-pi)

Answer 10

y - 3 = cos(π)(x-π)

Answer 11

yea but that's not what the book says -_-

Answer 12

Yes it is.

Answer 13

no.... lol i have the book in front of me? it says y = -x + pi +3 and the normal as y = x + pi +3 idk how they got that...

Answer 14

\[\large y - 3 = cos(π)(x-π) \rightarrow y = -x + π +3 \]

Answer 15

ok so the cosxpi is equivalent to -x?

Answer 16

cos(π) = -1, yes.

Answer 17

Remember you unit circle.

Answer 18

oh ok thanks