Real quick easy question: hori and vert tan lines

  • #1
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)
 
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Answers and Replies

  • #2
Dick
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You have a horizontal tangent where y'=0. You have a vertical tangent where y' goes to + or - infinity.
 
  • #3
tiny-tim
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Welcome to PF!

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

Hi ClaudeFrollo! Welcome to PF! :smile:

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

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. :wink:
 
  • #4
(1-x)/[tex]\sqrt{4-x^{2}}[/tex]=y'

y'= 1/2

Hmm... so does that mean that that's the slope of the horizontal tangent line?
 
  • #5
Dick
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(1-x)/[tex]\sqrt{4-x^{2}}[/tex]=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.
 
  • #6
Whoops. I think I meant:

x([tex]\sqrt{4-x^{2}}[/tex])/([tex]\sqrt{4-x^{2}}[/tex])

so... y'=-x? So x=0 when y'=0?
 
  • #7
Dick
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Whoops. I think I meant:

x([tex]\sqrt{4-x^{2}}[/tex])/([tex]\sqrt{4-x^{2}}[/tex])

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?
 
  • #8
tiny-tim
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Whoops. I think I meant:

x([tex]\sqrt{4-x^{2}}[/tex])/([tex]\sqrt{4-x^{2}}[/tex])

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

hmm :cry: … I think you need a Big Hint:

[tex]\sqrt{4\ -\ x^2}\ =\ \frac{4\ -\ x^2}{\sqrt{4\ -\ x^2}}[/tex]​
 
  • #9
Ohhhhh yeah, so I guess

((1-x)+(4-x[tex]^{2}[/tex]))/([tex]\sqrt{4-x^{2}}[/tex])

so f'(0)= 5/4?
 
  • #10
tiny-tim
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Ohhhhh yeah, so I guess

((1-x)+(4-x[tex]^{2}[/tex]))/([tex]\sqrt{4-x^{2}}[/tex])

so f'(0)= 5/4?

I'm confused … where did 1-x come from? :confused:
 
  • #11
I meant -x^2, not 1-x D:
 
  • #12
tiny-tim
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  • #14
Dick
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when x= -2, 2?

No. What did you finally get for the numerator of f'?
 
  • #15
tiny-tim
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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. :rolleyes:

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

Well? :smile:
 
  • #16
Well? :smile:

I thought I answered this when I said

((-x2(4-x2))/([tex]\sqrt{4-x^{2}}[/tex])

...wouldn't the numerator just be ((-x2(4-x2))?
 
  • #17
Dick
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I thought I answered this when I said

((-x2(4-x2))/([tex]\sqrt{4-x^{2}}[/tex])

...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?
 

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