Real quick easy question: hori and vert tan lines

[ClaudeFrollo]

sorry, i hit enter too soon

Homework Statement

Y=x(sqrt(4-x^2))

how do I find all pts on the graph where the tan line is vert and hori?

Homework Equations

Y=x(sqrt(4-x^2))

The Attempt at a Solution

y'=(x)(.5(4-x^2)^(-1/2)(-2x))+(sqrt(4-x^2))(1)

 

[Dick]

You have a horizontal tangent where y'=0. You have a vertical tangent where y' goes to + or - infinity.

 

[tiny-tim]

Welcome to PF!

y'=(x)(.5(4-x^2)^(-1/2)(-2x))+(sqrt(4-x^2))(1)

Hi ClaudeFrollo! Welcome to PF!

(have a square-root: √ and a squared: ² )

You have y' = -x²/√(4 - x²) + √(4 - x²),

and you want to solve y' = 0.

Hint: simplify the expression for y' by putting the whole thing as a fraction with √(4 - x²) on the bottom.

 

[ClaudeFrollo]

(1-x)/$$\sqrt{4-x^{2}}$$=y'

y'= 1/2

Hmm... so does that mean that that's the slope of the horizontal tangent line?

 

[Dick]

(1-x)/$$\sqrt{4-x^{2}}$$=y'

y'= 1/2

Hmm... so does that mean that that's the slope of the horizontal tangent line?

That's not y'. Try the algebra you used to simplify the derivative again. Then try and set y'=0 and figure out x and then the x values where y' goes to infinity.

 

[ClaudeFrollo]

Whoops. I think I meant:

x($$\sqrt{4-x^{2}}$$)/($$\sqrt{4-x^{2}}$$)

so... y'=-x? So x=0 when y'=0?

 

[Dick]

Whoops. I think I meant:

x($$\sqrt{4-x^{2}}$$)/($$\sqrt{4-x^{2}}$$)

so... y'=-x? So x=0 when y'=0?

No. I don't think you meant that either. Can you put the two terms from the product rule over a common denominator and add them?

 

[tiny-tim]

Whoops. I think I meant:

x($$\sqrt{4-x^{2}}$$)/($$\sqrt{4-x^{2}}$$)

so... y'=-x? So x=0 when y'=0?

hmm … I think you need a Big Hint:

$$\sqrt{4\ -\ x^2}\ =\ \frac{4\ -\ x^2}{\sqrt{4\ -\ x^2}}$$​

 

[ClaudeFrollo]

Ohhhhh yeah, so I guess

((1-x)+(4-x$$^{2}$$))/($$\sqrt{4-x^{2}}$$)

so f'(0)= 5/4?

 

[tiny-tim]

Ohhhhh yeah, so I guess

((1-x)+(4-x$$^{2}$$))/($$\sqrt{4-x^{2}}$$)

so f'(0)= 5/4?

I'm confused … where did 1-x come from?

 

[ClaudeFrollo]

I meant -x^2, not 1-x D:

 

[tiny-tim]

I meant -x^2, not 1-x D:

That's better!

So y' = 0 when … ?

 

[ClaudeFrollo]

when x= -2, 2?

 

[Dick]

when x= -2, 2?

No. What did you finally get for the numerator of f'?

 

[tiny-tim]

when x= -2, 2?

ClaudeFrollo, you'll make less mistakes if you always give a full answer, and don't try to do things in your head.

hmm … now seems as good a time as any to remark that your "Real quick easy question" was a clear breach of the Trade Descriptions Act ​

No. What did you finally get for the numerator of f'?

Well?

 

[ClaudeFrollo]

Well?

I thought I answered this when I said

((-x²(4-x²))/($$\sqrt{4-x^{2}}$$)

...wouldn't the numerator just be ((-x²(4-x²))?

 

[Dick]

I thought I answered this when I said

((-x²(4-x²))/($$\sqrt{4-x^{2}}$$)

...wouldn't the numerator just be ((-x²(4-x²))?

Don't you mean -x^2+(4-x^2)? While do you keep twisting things up?