Find the equation of y^2=x(x-3)^2 of tangent line at (3,0)

  • Thread starter Torshi
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  • #1
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Homework Statement



Find the equation of y^2=x(x-3)^2 of tangent line at (3,0)

Homework Equations


Given above.
I think implicit differentiation is involved or no since there is no xy's on the same side?


The Attempt at a Solution


Anyways...

My attempt:

2ydy/dx = x*2(x-3)*1
2ydy/dx=2x(x-3)
dy/dx= 2x(x-3)/2y
dy/dx= 2x^2-6x/2y


Doesn't seem right.
 

Answers and Replies

  • #2
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Another problem" Determine (x,y) location(s) where the graph of y^4 = y^2-x^2 has horizontal tangents

I got the answer dy/dx = -2x/4y^3-2y

I don't know how to calculate x and y positions? I just found the implicit differentiation
 
  • #3
35,115
6,856

Homework Statement



Find the equation of y^2=x(x-3)^2 of tangent line at (3,0)

Homework Equations


Given above.
I think implicit differentiation is involved or no since there is no xy's on the same side?


The Attempt at a Solution


Anyways...

My attempt:

2ydy/dx = x*2(x-3)*1
You have a mistake above. x * (x - 3)2 is a product, so the product rule is needed.

After you use the product rule, you will need to use the chain rule.
2ydy/dx=2x(x-3)
dy/dx= 2x(x-3)/2y
dy/dx= 2x^2-6x/2y


Doesn't seem right.
 

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