Question

Calculus: Need help on a tangent line question
Find the equation of the tangent line to the curve at the given point
x=cos(7y); ((-1/2)sqrt(3), (-1/6)pi)

Answer 1

So the curve is ((-1/2)sqrt(3), (-1/6)pi) ?

Do you know what the derivative of a curve equation gives you?

Answer 2

I got the derivative but im not sure if it is correct but

\[1=-7\sin7y \frac{ dy }{ dx }\]

Answer 3

I keep trying to plug to solve for dy dx but I end up with 1/-24.5

Answer 4

But the awnser choices are

Answer 5

Ok first of all I got that wrong. What I stated was the point. So for the equation, you'll have to solve for y and then derive. Let me see.

Answer 6

I get d/dx(1/7 cos^(-1)(x))

Answer 7

How did you get to that step?

Answer 8

DANGIT MY TYPING TODAY. What I put was the rearranged equation. Here lemme show you my steps from the beginning.

First I rearranged it in terms of y and get \[y=\frac{ \cos^{-1} x }{ 7 }\] Then I differentiate it. I afctor out the constant 1/7 and use the derivative of cos^(-1)(x) is -1/sqrt(1-x^2).

Answer 9

So the final derivative is \[\frac{ 1 }{ 7 }\frac{ -1 }{ \sqrt{1-x ^{2}} }\]

Answer 10

Ok so now I just plug in x and I get

\[(\frac{ 1 }{ 7 })(\frac{ -1 }{\sqrt{1-\frac{ \sqrt{3} }{ 2 }^2}}\]

Answer 11

I'm sure you can simplify that further. The square root of 3 is squared :)

Answer 12

is just 3/4

Answer 13

So the slope is (-4/7)

Answer 14

Hold up. Why does the picture you posted not look like the one we're doing here

Answer 15

Because it is the missed problem I got and I needed help on why I got it wrong

Answer 16

or am doing wrong

Answer 17

Well which one are you asking for help for right now? I've been doing the original one you posted. Also, you put in x wrong. Read again what given x you have. I think you just put in sqrt(3) instead of the whole thing.

Answer 18

They both are the same

Answer 19

They both are the same problem

Answer 20

Hmm. Well for the original one check again your work when pluggin in the x as I said. That will give you the slope. Now, for the second part, do you remember the slope of a horizontal line?

Answer 21

Yes you just set f(x)=0

Answer 22

correct?

Answer 23

or wait woops wrong problem on the picture

Answer 24

Not quite. You set f'(x)=0 It is the derivative, the slope that is 0

Answer 25

Lol sorry I dont need help on that one my bad

Answer 26

I uploaded the wrong one

Answer 27

Now im tripping lol

Answer 28

God my head hurts too now. Well let's continue where we left off then. Plug in x, correctly.

Answer 29

ok