Question

Please help with Calc problem!!!

Answer 1

I was able to find the first equation, but I am unsure how to find the other two. And then with letter b, I was just so lost! :(

Answer 2

did you take the derivative yet?

Answer 3

Yeah, I got it to be -2

Answer 4

This is calculus?

Answer 5

yup!

Answer 6

And for the derivative I got \[\frac{ 2x }{ 5y^2-1 }\] and when I substituted 1 in for x and 0 for y I got -2

Answer 7

Do you know how to get another tangent equation?

Answer 8

you can use the mean value theorem for A.
then (for b) point slope once you know at which values do tangent occur, and know the points

Answer 9

That derivative should be \[ \frac{2x}{5y^4-1}\]
Also, if you plug in x = 1, then the equation for the curve gives you
\[y^5 - y = 0 \rightarrow y(y^4-1) = y(y^2+1)(y^2-1) = y(y^2+1)(y+1)(y-1) = 0\]
From which you should notice that there are three possible values for y, leading to three tangent lines.

Answer 10

What is the mean value theorem?

Answer 11

I can't really explain that in one reply.... srry abt dat

Answer 12

go with Jemurray ...

Answer 13

Thank you Jemurray!! And then for part b, would the coordinates just be (1,0) (-1,0) and (0,0)??

Answer 14

The tangent line will be vertical when \(5y^4 - 1 = 0\), or \(y = \pm \frac{1}{\sqrt[4]{5}}\). The corresponding x values are
\[ x = \pm \sqrt{y(y^4-1) + 1} \]
So there will be four points where the tangent line is vertical.