Welcome to our community

Be a part of something great, join today!

Help with limits/tangent line

Status
Not open for further replies.

Jayden

New member
Jun 6, 2012
21
Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.



1. Can someone please explain what is going on in steps, really simply.
2. what is h?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.



1. Can someone please explain what is going on in steps, really simply.
2. what is h?
Hi Jaden,

The graph of the quadratic equation \(f(x)=x^2\) is shown in the picture. \(P\) is the point \((1,1)\) and \(Q\) is the point \(\left(1+h,(1+h)^2\right)\) where \(h\) is any real number. According to the picture \(h\) should be a positive number as well. Note that both of these points are on the graph since they satisfy \(f(x)=x^2\).

The gradient of the line that joins the two point \(P\mbox{ and }Q\) is given by, \(\displaystyle\frac{(1+h)^2-1}{(1+h)-1}=\frac{(1+h)^2-1}{h}\). Which is given as the slope of \(PQ\).

Now we shall generalize this for any two points.

Let, \(P\equiv(x_1,f(x_1))\mbox{ and }Q\equiv(x_1+h,f(x_1+h))\). Then the slope(say \(m_{PQ}\)) of the line \(PQ\) will be,

\[m_{PQ}=\frac{f(x_1+h)-f(x_1)}{(x_{1}+h)-x_1}=\frac{f(x_1+h)-f(x_1)}{h}\]

Hope this clarified all your doubts.
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.
Note that \(h\) is the separation between the two x-coordinates of the points \(P\) and \(Q\). When, \(h\rightarrow{0}\) the point \(Q\) will converge to \(P\). Ultimately you will have the tangent line. The gradient of the tangent(we shall denote this by \(f'(x_1)\)) is therefore given by,

\[f'(x_1)=\displaystyle\lim_{x\rightarrow 0}\frac{f(x_1+h)-f(x_1)}{h}\]

Now consider our function \(f(x)=x^2\). Try to substitute for \(f(x_1+h)\mbox{ and }f(x_1)\) in the above equation.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Re: Help with limits/tangaent line

Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.
The limit as \(h \to 0\) gives you the slope \( m \) of the tangent. You have found the slope to the curve at the point \((1,1)\), so the tangent is the line:

\( y=m x +c\)


and as it passes through \( (1,1) \) you have:

\(1=m + c\)

or

\(c=1-m\)

CB
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

Wait, so say we have f(x) = x^2

let a be the x value of point 1.

P1 = (a, f(a))
P2 = (a + h, f(a + h))

using the formula m = (y2 - y1) / (x2 - x1)

We get:

m = (f(a + h) - f(a)) / ((a + h) - a)

which then becomes:

m = (((a + h)^2) - (a^2)) / ((a + h) - a)

which is then simplified to:

m = (((a + h)^2) - (a^2)) / h

Am I right?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

Wait, so say we have f(x) = x^2

let a be the x value of point 1.

I hope you meant "Let a be the x value of point P1"

P1 = (a, f(a))
P2 = (a + h, f(a + h))

using the formula m = (y2 - y1) / (x2 - x1)

We get:

m = (f(a + h) - f(a)) / ((a + h) - a)

which then becomes:

m = (((a + h)^2) - (a^2)) / ((a + h) - a)

which is then simplified to:

m = (((a + h)^2) - (a^2)) / h

Am I right?
Yes, of course you are correct. So the slope of \(P_{1}P_{2}\) is \(\displaystyle m = \frac{(a + h)^2-a^2} {h}\). But this can be further simplified. Try to expand the square in the numerator.
 
Last edited:

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

(a^2 + 2ah + h^2) / h

?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

(a^2 + 2ah + h^2) / h

?
Incorrect. The original fraction is, \(\displaystyle m = \frac{(a + h)^2-a^2} {h}\). You have expanded \((a+h)^2\) but what about the \(-a^2\) term?
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

woops

2ah + h^2 / h
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

woops

2ah + h^2 / h
You can simplify further. Try to factor the numerator and cancel what is common to both the numerator and the denominator.
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

2ah + h^2 / h

h(2a + h) / h

h(2a) / 1

2ah / 1

2ah?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Re: Help with limits/tangaent line

2ah + h^2 / h

h(2a + h) / h

h(2a) / 1

2ah / 1

2ah?
Almost.

\(\displaystyle \frac{2ah+h^2}{h}\)

\(\displaystyle \frac{h \left(2a+h \right)}{h}\)

\(\displaystyle 2a+h\)

So the h in the numerator and denominator cancel and the above line is what you are left with.
 

checkittwice

Member
Apr 3, 2012
37
Re: Help with limits/tangaent line

woops

2ah + h^2 / h

You must use grouping symbols around the numerator to be correct:


(2ah + h^2)/h


Otherwise, what you typed is equivalent to:

[tex]2ah \ + \ \dfrac{h^2}{h}.[/tex]
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

Howcome you cancel the h outside the brackets instead of the other h? Is that because the other h is inside brackets and therefore attached to the 2a?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Help with limits/tangaent line

Howcome you cancel the h outside the brackets instead of the other h? Is that because the other h is inside brackets and therefore attached to the 2a?
Hi Jayden,

Brackets are used to specify the order of operations. If you write, \(h(2a+h)\), this means that you have to add \(2a\mbox{ and }h\) first and then multiply it with \(h\). You can consider, the things which are inside the parenthesis as one unit. Now you have, \(\dfrac{h(2a+h)}{h}\), you have to multiply \(h\) with \(2a+h\) and divide the answer by \(h\). Multiplication and division have the same order of precedence and therefore you can divide the \(h\) in the numerator by the one in the denominator.

To refresh yourself about the order of operations I kindly suggest you to read, this and this.

Kind Regards,
Sudharaka.
 

Jayden

New member
Jun 6, 2012
21
Re: Help with limits/tangaent line

I have a question. This one always gets me.

Say we have \(\dfrac{9x^2 + 5}{3x^2}\), does that make it \(\dfrac{6x^2 + 5}{1}\)

Also if we have \(\dfrac{6x^2 + 5}{18x^2}\), what would be the simplified answer? This got me in my last test.



EDIT: uhh how do you display fractions in that font?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Re: Help with limits/tangaent line

You might need some time to learn the Latex syntax. We have a forum here with good information on how to get started. Also, this is a new topic so please start a new thread. We are more than happy to help you and very happy you are here! :)
 
Status
Not open for further replies.