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

In summary, the problem involves finding the equation of the tangent line to the curve y^2=x(x-3)^2 at the point (3,0). Implicit differentiation is used to solve the problem. After using the product rule and chain rule, the derivative of y with respect to x is found to be 2x^2-6x/2y. However, it is noted that this answer may not be correct.
  • #1
Torshi
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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.
 
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  • #2
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
Torshi said:

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.
Torshi said:
2ydy/dx=2x(x-3)
dy/dx= 2x(x-3)/2y
dy/dx= 2x^2-6x/2y


Doesn't seem right.
 

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

1. What is the equation of y^2=x(x-3)^2?

The equation y^2=x(x-3)^2 represents a parabola that opens to the right. It is a degree 4 polynomial equation.

2. How do you find the tangent line at a given point on the curve?

To find the tangent line at a given point on the curve, we first find the derivative of the equation using the power rule. Then, we substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. Finally, we use the point-slope form of a line to find the equation of the tangent line.

3. What is the significance of the point (3,0) in this equation?

The point (3,0) represents a point of tangency on the curve y^2=x(x-3)^2. This means that the tangent line at this point is parallel to the curve and touches the curve at only one point.

4. Can you use the equation of the tangent line to approximate values of y on the curve?

Yes, the equation of the tangent line can be used to approximate values of y on the curve at points close to the point of tangency. However, the further away from the point of tangency, the less accurate the approximation will be.

5. How can you determine if a point lies on the tangent line of a curve?

A point (x,y) lies on the tangent line of a curve if it satisfies the equation of the tangent line. This means that the slope of the tangent line at that point is equal to the slope of the line connecting the given point and the point of tangency on the curve.

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