Math Is Fun Forum

Denominator

Member

Registered: 2009-11-23

Posts: 220

Tangent Help!

Hey guys,
is there a formula to figuring out tangents of let say f(x) if only a point of the tangent (a, b), (not on f(x)) is given?

Thanks a bunch guys

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tangent Help!

Hi Denominator;

That is not always possible.

Take a look at the picture. See the point near (1,6)? I don't think it is possible to draw a tangent fronm it to any point on y = x^2.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Denominator

Member

Registered: 2009-11-23

Posts: 220

Re: Tangent Help!

Okay but if its a function like y=x^2 something like that but not like functions with gaps or multiple values, if u know w hat i mean

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tangent Help!

Hi Denominator;

Sometimes it can be done. If your point was (2,1) then it could be done.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Denominator

Member

Registered: 2009-11-23

Posts: 220

Re: Tangent Help!

o.o i gotta tell my maths teacher that!
But if it is possible for a tangent to go on a point, is there like a formula to figure that out, or do i have to guess and check?

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tangent Help!

I am not sure, since it doesn't hold for all possible points, it might not be possible to have a strict formula. I will work on it and post here if I find something.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Denominator

Member

Registered: 2009-11-23

Posts: 220

Re: Tangent Help!

Thank you bobbym this will be much appriciated

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tangent Help!

Hi;

Here is a general formula, I have developed for this problem. Don't get too excited over it, it has not been tested exhaustively. If we have

y = f(x) and y - y1 = m(x - x1)

m = f '(x)

y - y1 = f '(x) ( x - x1)

y = f '(x) (x - x1) + y1

f(x) = y = f '(x) (x - x1) + y1

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Denominator

Member

Registered: 2009-11-23

Posts: 220

Re: Tangent Help!

omg thank you very much!
so is x1 and y1 the point?

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tangent Help!

Yes:

Let's see how this works. We have the point (2,1) and y = x^2 we want the tangent to the curve through that point.

Using the general formulas derived in post #8.

y = x^2 and y - 1 = 2x(x - 2)

y = 2x(x-2) +1

x^2 = 2x(x-2)+2

Solve for x, x = 2 + √3 and x = 2 - √3

Do you see what you have to do now?

(2,1) is one point, the one you are given. x = 2 +√3 is another , it must lie on the curve so y = f(2 +√3)=7 + 4√3

Now you have 2 points (2,1) and (2 +√3,7 + 4√3) these determine the tangent that passes through (2,1) and touches y = x^2

Now use the equation through 2 point formula:

or

That is the equation of the tangent of y = x^2 through (2,1).

I found this page:

http://www.karlscalculus.org/calc4_5.html

It seems I am a little late with my idea. Karl got there first!

Also there is another tangent that corresponds to x = 2 - √3  on the curve y = x^2.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.