Question

Find the equation of the line that is tangent to the function y=-4sinx at x= pi/4

Answer 1

\[\huge x=\frac{ \pi }{ 4 } \]

Answer 2

\[\huge y'= -2\sqrt{2}\]

Answer 3

There are two steps you need to complete here. First find the derivative, and then plug in the value of x into that derivative. Do you know how to take the derivative?

Answer 4

Wait why is the derivative -2(sqrt2)?

Answer 5

\[\huge y'(\frac{ \pi }{ 4 })\]

Answer 6

I forgot to but the brackets

Answer 7

Oh okay. Well just do as I said originally and find the derivative. Do you know what it is?

Answer 8

\[\huge y'= -4\cos(\frac{ \pi }{ 4 })\]

Answer 9

So the slope of the tangent line is going to the value you get when you plug in x. You already seemed to have gotten it, so that's the first part. I'm actually blanking out on finding the y intercept of that line... I apologize, it's been a while -.-

Answer 10

It's okay lol :P

Answer 11

find the derivative of the y function and then plug in the x point to find the slope of the tangent line

Answer 12

i did

Answer 13

i need the equation though @yummydum

Answer 14

You have the slope and a point. How about the point-slope formula?\[y-y_1=m(x-x_1)\]

Answer 15

and your \[m=y'=-2\sqrt{2}\] so all you have to do is use your coordinates and your slope, or \[y'\]

Answer 16

once you get the derivative, plug in the x value, that gives you the m, then plug in the x value into the original equation and that gives you the y

so then you will have x, m, and y, and you can fill in the point slope formula

Answer 17

Texrbook says: \[y= -2\sqrt{x}+\frac{ \sqrt{2\pi} }{ 2 }-2\sqrt{2}\]

Answer 18

That's when you get all your values and plug them in, you will get that as your result.

Answer 19

im not getting that tho ..

Answer 20

I dont get where the 1st term is coming from

Answer 21

I see your point on that. I would get \(-2x\sqrt{2}\)

Answer 22

They did something similar with the pi.

Answer 23

y=-4 sinx
\[when x= \frac{ \pi }{ 4 } ,y=-4\sin \frac{ \pi }{ 4 },y=-4\times \frac{ 1 }{ \sqrt{2} }=\frac{ -4\sqrt{2} }{2}=-2\sqrt{2}\]
dy/dx=-4cos x
\[at x=\frac{ \pi }{ 4 },\frac{ dy }{ dx }=-4\cos \frac{ \pi }{ 4 }=-4*\frac{ 1 }{ \sqrt{2} }=-2\sqrt{2}\]
\[eq. of tangent line is y-\left( -2\sqrt{2} \right)=-2\sqrt{2}\left( x-\frac{ \pi }{4} \right)\]

Answer 24

simplify & get the reqd. eq

Answer 25

oops, yah, went back to the original, not the derivative.

Answer 26

hmmm... but still, how would the x and \(\pi\) end up inside the roots to get the book answer.

Answer 27

Shouldn't the slope be \[-2\sqrt{2}\]? His textbook answer is only stating -2, disregarding the rad2. How is that possible?

Answer 28

\[y=-2\sqrt{x}+\frac{ \sqrt{2\pi} }{ 2 }-2\sqrt{2}\] is his answer, but i think it should be \[y=-2\sqrt{2}x+\frac{ \sqrt{2\pi} }{2 }-2\sqrt{2}\]

Answer 29

and how i got this is by following these steps: \[y=-4sinx\] when \[x=\frac{ \pi }{ 4 }\] So you have to take the derivative to find the slope first. \[y' = -4cosx\] and then you can plug in yourx-value to find y'\[y' = -4\cos(\frac{ \pi }{ 4 }) = -4(\frac{ \sqrt{2} }{ 2 })= -2\sqrt{2}\] And then to find the y-intercept, plug the x-value given into the equation:\[y=-4\sin (\frac{ \pi }{ 4 })= -4(\frac{ \sqrt{2} }{ 2 })=-2\sqrt{2}\] Now you have your slope, the y=-intercept, and your x-intercept. You can use the point slope form to evaluate your equation: \[y-y _{1}=m(x-x _{1})\] INputting values, you get:\[y-(-2\sqrt{2})=-2\sqrt{2}(x-\frac{ \pi }{ 4 })\]\[y+2\sqrt{2}=-2\sqrt{2}x+\frac{ 2\sqrt{2\pi} }{ 4}\]\[y=-2\sqrt{2}x+\frac{ \sqrt{2\pi} }{ 2 }-2\sqrt{2}\]

Answer 30

I hope it helps somewhat :\

Answer 31

pi is not in the square root

Answer 32

is it just \[\frac{ \sqrt{2} }{ 2 }\pi\] ?

Answer 33

Exacty, so either he has copied his textbook incorrectly or there is an error in his textbook.