Hi there,To find the gradient of a curve, we draw a chord on the

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Hi there,

To find the gradient of a curve, we draw a chord on the curve and then makes the 2nd coordinates ( B ) tends to A ( 1st coordinates ).

To find the gradient of the chord, i.e, ΔY/ΔX, we replace the two coordinates into the equation of the curve. But my question is why do we replace them ?
 

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  • #2
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The gradient field evaluated along a curve gives you vectors normal to the curve. If you draw a chord (hence, a straight line), its gradient will be lines (or, rather vectors along these lines) of the form y = ax +b
 
  • #3
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Sorry, but I didn't quite understood there. Can you explain more simpler please. Sorry again.
 
  • #4
HallsofIvy
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I think meldraft is misunderstanding your use of the word "gradient". In the United States, we typically just use the word "gradient" to mean [itex]\nabla f= \partial f/\partial x\vec{i}+ \partial f/\partial y\vec{j}+ \partial f/\partial z\vec{k}[/itex], which will give a vector normal to the curve f(x,y,z)= constant.

But in the United Kindom, and other places, "gradient" is used to mean just "derivative".
And that is defined as the limit of the "difference quotient":
[tex]\frac{f(x)- f(x_0)}{x- x_0}= \frac{y(x)- y(x_0)}{x- x_0}[/tex]

Geometrically that "difference quotient" is the slope of the line from [itex](x_0, y_0)[/itex] to (x, y) so the limit gives the slope of the tangent line.
 
  • #5
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Ahhh I didn't know that, thanks HallsofIvy!

So, if I understand correctly, your question is why we evaluate the formula that HallsofIvy wrote, with points A and B.

As HallsofIvy explained, the limit gives you the tangent line.

In other words, if you don't evaluate the formula, you won't get a number for the slope (derivative) of the curve at this point.
 
  • #6
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Yeah by gradient, I meant the slope of a line. My question is why do we replace the 2 coordinates of the line into the equation of the curve to find the gradient of the line ?

And ultimately move the line and make it become a tangent to the curve so as to find the gradient of the curve.

Just to say that, I'm still at a low level and have just started learning Differentiation and I don't know anything about difference quotient.

Searched for difference quotient in Wikipedia and I got to know that its related to function and difference of 2 points. The rest I didn't understood.
 
  • #7
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The concept is that you split the curve into infinitesimally small lines, each one after the other, and each with a different slope. The equation that gives you the slope of a line is:

[tex]\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_1+Δx)-f(x_1)}{Δx}[/tex]

where points 1,2 are the beginning and the end of the line, respectively (or, in principle, any 2 points along the line)

Now, if you do this for the entire curve, you get its derivative:

[tex]f'(x)=lim_{Δx->0} \frac{f(x+Δx)-f(x)}{Δx}[/tex]
 
  • #9
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Is this from a numerical analysis/computational mechanics course?

What that example is probably demonstrating, is that by calculating f(x) and f(Δx) and replacing in the derivative formula, you get the general formula for the actual derivative. This happens because Δx can be neglected from the result, since it's infinitely small (therefore its length is practically 0). Hence, you have proven that the derivative of x^2 is 2x :)

Edit: as for your question, no, if you do it by hand you do not actually calculate every single segment (although, theoretically that is the concept). By calculating that limit you get a new function that basically describes how the slope changes, as x grows larger/smaller (that function is the derivative of the curve)
 
  • #10
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by calculating f(x) and f(Δx) and replacing in the derivative formula,
What derivative formula ? From the example, f(x) and f(Δx) were replaced in the equation of the curve which in this case is y=x^2.

My original question was why do we replace f(x) and f(Δx) in the equation. What does it do ?

For me its a kind of a simultaneous equation and from what I know is that we do simultaneous equation to know at which coordinate(s) two lines meet.
 
  • #11
HallsofIvy
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First, we don't "replace [itex]f(\Delta x)[/itex]", we replace [itex]f(x+ \Delta x)[/itex]. Now, apparently, I don't understand your qestion. We replace f(x) with [itex]x^2[/itex] because we are asked to find the derivative of the function [itex]f(x)= x^2[/itex]. And, we replace [itex]f(x+\Delta x)[/itex] with [itex](x+\Delta x)^2[/itex] because that is what [itex]f(x+\Delta x)[/itex] means!

Is the difficulty that you do not understand "functional notation"? If we were given, instead, that [itex]f(x)= x^3[/itex], then we would know that [itex]f(2)= 2^3= 8[/itex], [itex]f(a)= a^3[/itex], [itex]f(x+1)= (x+ 1)^3= x^3+ 3x^2+ 3x+ 1[/itex], and that [itex]f(x+ \Delta x)= (x+ \Delta x)^3= x^3+ 3x^2\Delta x+ 3x(\Delta x)^2+ (\Delta x)^3[/itex]. Then, of course, [itex]f(x+\Delta x)- f(x)= x^3+ 3x^2\Delta x+ 3x(\Delta x)^2+ (\Delta x)^3- x^3= \Delta x(3x^2+ 3x\Delta x+ (\Delta x)^2)[/itex]
That means
[tex]\frac{f(x+\Delta x)- f(x)}{\Delta x}= \frac{\Delta x(3x^2+ 3x\Delta x+ (\Delta x)^2)}{\Delta x}= 3x^2+ 3x\Delta x+ (\Delta x)^2[/tex]
Now, what is the limit of that as [itex]\Delta x[/itex] goes to 0? What is the derivative of [itex]f(x)= x^3[/itex]?
 
  • #12
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First, we don't "replace [itex]f(\Delta x)[/itex]", we replace [itex]f(x+ \Delta x)[/itex]
Sorry, typo :smile:
 
  • #13
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The answer is 3X2

Edit : I do know how to work it out. But I don't know what each step does and I'm trying to know why. Its like I learned it by heart what to do from my school teacher and I know that's wrong.
 
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  • #14
HallsofIvy
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The basic idea is that you are looking at the slope of the line from the point [itex](x_0, f(x_0))[/itex] to the "neighboring" point [itex](x_0, f(x_0+ \Delta x))[/itex] (a "secant line" because a "secant line" in a circle is two a line that passes through distinct points on the circle while a "tangent line" only touches a single point on the circle). The reason we replace [itex]x_0[/itex] with [itex]x_0+ \Delta x[/itex] is that it moves us over a little bit so that we do have two points and so can calculate a slope. Then take the limit as [itex]\Delta x[/itex] goes to 0. (Not "until ΔX becomes 0"- take the limit. If you don't have a pretty good grasp of limits it will be hard to understand the derivative.) This can be done because, geometrically, the limit of those "secant lines" is the "tangent line" to the curve at that point.

This has, by the way, nothing to do with "differential equations" so I am moving it to "Calculus and Analysis".
 
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  • #15
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The reason we replace [itex]x_0[/itex] with [itex]x_0+ \Delta x[/itex] is that it moves us over a little bit so that we do have two points and so can calculate a slope.
You mean to calculate the slope of the tangent ?

So let me say what I've understood :

At first we were at the point [itex]x_0[/itex], f([itex]x_0[/itex]) and by replacing the neighbouring coordinate, we moved a little bit and now we see the initial coordinate( where I was ) and the neighbouring coordinate. So we can calculate the slope of the secant line ( gradient ).

Am I right here ?

Edit : I've checked Wikipedia and it gives the definition of limit used in Mathematics. But do you know somewhere that I can have deeper understanding for it ?
 
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