Implicit vs non implicit diff. (line passing through point & tangent to function)

In summary, the conversation discusses finding equations of lines that pass through a given point and are tangent to a given curve. The confusion lies in the use of implicit differentiation and the difference between implicit and non-implicit problems. The equation of the line y = f'(x)(x) does not work for implicit problems because y is not a function of x. This can be solved by either getting everything in terms of x before taking the derivative, or by using substitution to solve for x. Once x and y values are found, the slope and equation of the line can be determined.
  • #1
LearninDaMath
295
0

Homework Statement




Find the equations of the lines that pass through (0,0) and are tangent to x^2 - 4x + y^2 + 1 = 0


My confusion

I've been given a problem of this sort recently, except now it involves implicit differentiation. I know "how" to get to the correct answer. I just don't "understand" what's going on as I'm getting to the correct answer.

The problem I had recently was: Find the equations of the lines that pass through (0,0) and are tangent to "f(x) = x^3 - 9x^2 - 16x = y

When comparing these two problems, the process of finding the answer seems to be different, however, the question is essentially asking the same exact thing in both problems. So I must not be seeing "how" the process of solving each is actually similar or the same.

Here is the first one that I solved:

implicita.jpg



However, when I follow the same exact process for the current problem involving implicit differentiation, I get this:


implicitc.jpg




In the first problem, I am supposed to be looking for the equation of a line: (y-y)=m(x-x) and it becomes f(x) = (f'(x))(x)

However, the case of implicit differentiation does not seem to want to follow the same process, which leads me to the observation that:

in the first non implicit problem, f(x) = x^3 - 9x^2 - 16x fits neatly into the left side of equation y=f'x(x)

while

in the implicit problem f(x) = x^2 - 4x + y^2 + 1 = 0 does not fit neatly into the left side of equation y=f'x(x)

Why does f(x) = x^2 - 4x + y^2 + 1 = 0 not fit neatly into the left side of equation y = f'x(x) ? I thought that f(x) always equals y, since the equation for a function seems to always be expressed as "f(x) = y"
-----------------------------

Also, in the non implicit problem, f'(x) = 3x^2 - 18x - 16 seems to fit neatly into the m term of y=f'x(x)

while

in the implicit problem, f'(x) = 2x - 4 + 2y(dy/dx) = 0 does not fit neatly into the m term of y = f'x(x)

Why does f'(x) = 2x - 4 + 2y(dy/dx) = 0 not fit neatly in m? I thought that f'(x) was supposed to represent the slope of the tangent, thus fitting neatly into the m term (since it works that way in the non implicit problem).
------------------------------

Is it the fact that the equation is set equal to zero in the implicit problem while the function in the first problem is not set equal to zero? Does that have any impact on the process of solving this problem, if so, what impact does the zero have?

Is it the fact that there are x's and y's together in the implicit problem that is the factor that doesn't allow the entire f'x to fit neatly into the m term?

Why must we change the functions in order to arrive at the correct x value?
 
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  • #2


You write, repeatedly, "f(x)" while there is no such "f". y is NOT a function of x.
 
  • #3


HallsofIvy said:
You write, repeatedly, "f(x)" while there is no such "f". y is NOT a function of x.

This was a very confusing problem as it is. But the fact that I didn't understand that if an equation is in terms of both x and y's on one side, then it can't be considered a function of the form "f(x) = y. Thus its no wonder why it does fit so neatly. I would either have to get everything in terms of x before taking the derivative...or, as an easier route, i could do some substitution to solve for x. Once I solve for x, I then need to find y. Once I have both x and y values, I can then find the slope value. Then, the equation of the line(s) can be formed.

Thanks
 

What is the difference between implicit and non-implicit differentiation?

Implicit differentiation is a method for finding the derivative of a function that is not explicitly written in the form y = f(x). Non-implicit differentiation, also known as explicit differentiation, is used for functions that are already written in the form y = f(x).

When is implicit differentiation used?

Implicit differentiation is used when it is not possible or convenient to express the dependent variable, y, as a function of the independent variable, x. This often occurs when the function is defined implicitly, such as in the case of a circle or ellipse.

How is implicit differentiation performed?

To perform implicit differentiation, the derivative of both sides of the equation is taken with respect to the independent variable. The chain rule is then used to differentiate any terms that involve the dependent variable, treating it as a function of the independent variable.

What is the advantage of using implicit differentiation?

Implicit differentiation can be useful when finding the derivative of a function would be difficult or impossible using explicit differentiation. It allows us to find the derivative of functions that are not easily expressed in terms of y = f(x).

How does implicit differentiation relate to tangents and lines passing through points?

Implicit differentiation can be used to find the equation of a tangent line to a curve at a given point. This is done by finding the derivative of the function implicitly and then plugging in the coordinates of the point to find the slope of the tangent line. The equation of the tangent line can then be written in point-slope form using the point and slope found from the derivative.

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