Question

Find an equation of the tangent line to y = 2 sin x at x = π/6

Answer 1

take the derivative, get
\[2\cos(x)\] then replace x by
\[\frac{\pi}{6}\]

Answer 2

youre answer is in like slashes and [][ so it looks like \2\xos(x)\] \frac{pi6\]

Answer 3

so what is the answer to it?

Answer 4

He said to take the derivative and get 2cos(x).
Then replace x with pi over 6

Answer 5

my homework said 2cos(pi/6) was wrong

Answer 6

When you take the derivative, you get the slope function. When you replace x with pi over 6 you are finding the slope of the tangent at pi/6. Then you must use that slope to find the equation of the tangent line which the problem requires.

Answer 7

so is 2cos(pi/6) the answer or does it have more to it?

Answer 8

What are you supposed to find?

Answer 9

Find an equation of the tangent line to y = 2 sin x at x = π/6

Answer 10

Do you know what the equation of a line looks like?

Answer 11

y = mx+b?

Answer 12

yes. So I don't think 2cos(pi/6) looks much like the equation of a line, do you?

Answer 13

nope, pi/6x + 2cos? or the other way around?

Answer 14

Can you evaluate 2cos(pi/6)?

Answer 15

i have no idea?

Answer 16

\[2\cos \frac{\pi}{6}=2\times \frac{\sqrt{3}}{2}=\sqrt{3}\]

Answer 17

Does that look familiar?

Answer 18

Have you ever studied the special right triangles or the unit circle?

Answer 19

it has lines and stuff in it... does it say 2cos(pi/6) = 2 * sqrt3/2

Answer 20

Yes. And that 2(sqrt3/2)=sqrt3

Answer 21

could u help me pls after this?

Answer 22

What do you need help with?

Answer 23

http://openstudy.com/updates/4f4117b7e4b0534c53cc4ae7

Answer 24

so in all the equation is?

Answer 25

So sqrt3 is the slope of the tangent line.

Answer 26

And it goes through a point on the graph of y = 2sin(x) that has an x coordinate of pi/6

Answer 27

so sqrt3x+cos(pi/6)

Answer 28

If pi/6 is the x value, then 1 is the y value. So now we know the slope of the line and a point on the line. We should be able to write the equation using point slope form.

Answer 29

y-1=sqrt3(x-pi/6)

Answer 30

And that is the equation of the tangent line.

Answer 31

this is way too much work for a problem lol, so its y - 1 = sqrt3(x-pi/6)

Answer 32

It's hardly any work at all if you know what you are doing.

Answer 33

the homework said it was wrong :\

Answer 34

y = 2sinx --> f(pi/6) = 2sin(pi/6) = 1
y' = 2cosx
At x = pi/6, y' = 2cos(pi/6) = sqrt(3)
Tangent line:
y - 1 = V3 ( x - pi/6)
=>y = V3x - piV3/6 + 1

Answer 35

Then you should double check, otherwise I'm sure your homework solution's misprinted!

Answer 36

so is it y = sqrt3x - (pisqrt3/6) + 1

Answer 37

In short form, it is: y - 1 = V3 ( x - pi/6)

Answer 38

Most of the time, people want slope y intercept form, so you need to multiply it out.

Answer 39

so is that how i should put it in? how i did it?

Answer 40

Since tangent line is straight line:
y = mx + b

Answer 41

you need to find slope m!

Answer 42

m = derivative of function at specific point!

Answer 43

If you still have question, just let know!

Answer 44

i got it wrong

Answer 45

How, show me!

Answer 46

idk, i put it in like that and it was my fifth try so it said i was wrong

Answer 47

idk, i put it in like that and it was my fifth try so it said i was wrong

Answer 48

Why don't you try to V3 = 1.73, pi = 3/14, calculate yourself
My approximation is y = 1.73x + .0931