# Equation of tangent

• June 22nd 2010, 10:57 AM
bobsanchez
Equation of tangent
I understand how when doing the equation of a tangent line, you are given the equation and point P, you choose point Q, and from there you set up the limit.

My question is: how, exactly, do you pick point Q and set it all up exactly? My book is incredibly vague on this section. Everything else it is pretty explicit about, but this one has an example that doesn't make much sense and no real explanation of it.
• June 22nd 2010, 11:19 AM
Hello bobsanchez
Quote:

Originally Posted by bobsanchez
I understand how when doing the equation of a tangent line, you are given the equation and point P, you choose point Q, and from there you set up the limit.

My question is: how, exactly, do you pick point Q and set it all up exactly? My book is incredibly vague on this section. Everything else it is pretty explicit about, but this one has an example that doesn't make much sense and no real explanation of it.

The usual method is to let the coordinates of the point P be $(x,y)$ and Q be a neighbouring point at $(x+\delta x, y+\delta y)$. Then the gradient of the line PQ is:
$\dfrac{\text{increase in }y}{\text{increase in }x}$
$=\dfrac{\delta y}{\delta x}$
You then:
• Use the relationship between $x$ and $y$ (which you will have been given) to write down an expression for $\delta y$ in terms of $x$ and $\delta x$

• Eliminate $\delta y$ from the expression for the gradient, by using this relationship

• Simplify the result, and hence find the limit of this expression as $\delta x \to 0$.

For example, if we are told that the relationship between $x$ and $y$ is:
$y = x^2$
Then
$y+ \delta y = (x+\delta x)^2$ (because the coordinates of Q also satisfy the relationship)
$=x^2+2x\delta x + \delta x^2$
$\Rightarrow \delta y = 2x\delta x + \delta x^2$
So the gradient of PQ is $\dfrac{\delta y}{\delta x}$
$=\dfrac{2x\delta x + \delta x^2}{\delta x}$

$=2x + \delta x$

The limit of this expression as $\delta x \to 0$ is then clearly $2x$.

• June 22nd 2010, 11:19 AM
ebaines
Can you be more specific about this - what exactly do you mean by a point Q? Pehaps if you posted the example that is in your book we could explain it better.
• June 23rd 2010, 08:55 AM
bobsanchez
So basically add h to whatever the quantity is and then subtract that quantity from it?
• June 23rd 2010, 08:58 AM
bobsanchez
Quote:

Originally Posted by ebaines
Can you be more specific about this - what exactly do you mean by a point Q? Pehaps if you posted the example that is in your book we could explain it better.

Point Q is what the book called the coordinate you pick to subtract P from to get the equation. The example posted above is actually very similar to the one in my book.
• June 23rd 2010, 11:22 PM
Hello bobsanchez
Quote:

Originally Posted by bobsanchez
So basically add h to whatever the quantity is and then subtract that quantity from it?

Yes - more or less.

An alternative notation to $\delta x,\;y,\;\delta y$ is to use function notation - $y=f(x)$ - so that P is the point $\big(x,f(x)\big)$ and Q is $\big(x+h,f(x+h)\big)$. The gradient of PQ is then
$\dfrac{f(x+h)-f(x)}{h}$
This looks quite different from what I wrote in my first reply, but is simply an alternative way of saying the same thing.

In the example I gave you before, then, $f(x) = x^2$, and so the gradient of PQ is
$\dfrac{f(x+h)-f(x)}{h}=\dfrac{(x+h)^2-x^2}{h}$
$=\dfrac{x^2+2xh+h^2-x^2}{h}$

$=\dfrac{2xh+h^2}{h}$

$=2x+h$
So, as before, as $h\to 0$ the limit of this gradient - which is therefore the gradient of the tangent at P - is $2x$.