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A basic calculus question?

by zeromodz
Tags: basic, calculus
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zeromodz
#1
Jul29-10, 11:44 AM
P: 239
Okay, I am new to calculus so can you guys just explain something to me.

What is exactly are you doing to a function when you differentiate or take the derivative of it? I thought that when you do that, you make an instantaneous solution to where something is at all times with respect to the other axis. Is that correct?

Also, whenever you see calculus in advanced physics, I don't understand. You can only use indefinite integrals and derivatives when you have a valid function right? So all these advanced physics problems need functions to go along with them right?

Thank you guys!
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arildno
#2
Jul29-10, 12:19 PM
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What is exactly are you doing to a function when you differentiate or take the derivative of it? I thought that when you do that, you make an instantaneous solution to where something is at all times with respect to the other axis. Is that correct?
Hmm..

Try to formulate what the following quantity is, for non-zero h:
[tex]\frac{f(x+h)-f(x)}{h}[/tex]

Hint:
Draw an arbitray graph, pick two points on the x-axis ("x" and "x+h"), and think of h=(x+h)-x
zeromodz
#3
Jul29-10, 12:23 PM
P: 239
Quote Quote by arildno View Post
Hmm..

Try to formulate what the following quantity is, for non-zero h:
[tex]\frac{f(x+h)-f(x)}{h}[/tex]

Hint:
Draw an arbitray graph, pick two points on the x-axis ("x" and "x+h"), and think of h=(x+h)-x
This formula gives you the difference of Y over X. In other words, it gives you a tangent line over the 2 points you select. Right? Thats what it does.

Mark44
#4
Jul29-10, 12:24 PM
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A basic calculus question?

The derivative of a function is another function that gives the slope of the tangent line to the graph of the first function.
Quote Quote by zeromodz
Also, whenever you see calculus in advanced physics, I don't understand. You can only use indefinite integrals and derivatives when you have a valid function right? So all these advanced physics problems need functions to go along with them right?
Well of course. In order to differentiate a function or find the antiderivative of one, you need a function to operate on.
zeromodz
#5
Jul29-10, 12:27 PM
P: 239
Quote Quote by Mark44 View Post
The derivative of a function is another function that gives the slope of the tangent line to the graph of the first function.
Well of course. In order to differentiate a function or find the antiderivative of one, you need a function to operate on.
Okay thank you so much. I understand this a lot better now.
arildno
#6
Jul29-10, 12:48 PM
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Quote Quote by zeromodz View Post
This formula gives you the difference of Y over X.
Yes. And?
In other words, it gives you a tangent line over the 2 points you select. Right?
No, it does not.

a) First off, what do you call a straight line going through two points on the graph?

b) What, EXACTLY, is the QUANTITY you calculate?
A line is not a "quantity" in this sense!
zeromodz
#7
Jul29-10, 12:50 PM
P: 239
Quote Quote by arildno View Post
No, it does not.

a) First off, what do you call a straight line going through two points on the graph?

b) What, EXACTLY, is the QUANTITY you calculate?
A line is not a "quantity" in this sense!
a)A secant line! sorry, I just knew that I just put tangent line down for some reason.


b) You are calculating the area under the curve????
arildno
#8
Jul29-10, 12:53 PM
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Quote Quote by zeromodz View Post
a)A secant line!
Yes it is!
b) You are calculating the area under the curve????
Is the area under the curve equal to the ratio between the difference of Y-values and X-values??

If not, what geometric interpretation does this ratio have?
zeromodz
#9
Jul29-10, 12:58 PM
P: 239
Quote Quote by arildno View Post
Yes it is!

Is the area under the curve equal to the ratio between the difference of Y-values and X-values??

If not, what geometric interpretation does this ratio have?
The ratio of Y over X is tangent. Since the derivative of tangent is sec^2, that is why i get a secant line right? Also, the formula [tex]\frac{f(x+h)-f(x)}{h}[/tex] is what gives you the area under the curve? Sorry I am evading your question with another, just trying to clear things up.
arildno
#10
Jul29-10, 01:05 PM
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Quote Quote by zeromodz View Post
The ratio of Y over X is tangent.
Ok.
Now I get some of your problems.

No, the calculated number will not be "tangent" to the curve here, but it WILL give you the tangent value associated with the angle formed by the secant line and the horizontal line formed at the point (x,f(x)).

You have a natural triangle here in the (x,y)-plane, with corners:
(x,f(x)), (x+h,f(x)) and (x+h,f(x+h))

We usually call that ratio the slope of the line.

Have you heard that expression?
zeromodz
#11
Jul29-10, 01:12 PM
P: 239
Quote Quote by arildno View Post
Ok.
Now I get some of your problems.

No, the calculated number will not be "tangent" to the curve here, but it WILL give you the tangent value associated with the angle formed by the secant line and the horizontal line formed at the point (x,f(x)).

You have a natural triangle here in the (x,y)-plane, with corners:
(x,f(x)), (x+h,f(x)) and (x+h,f(x+h))

We usually call that ratio the slope of the line.

Have you heard that expression?
No, sorry. Its just whenever I thought of Tangent, I thought it was the slope which was rise over run (Y / X). I clearly don't have much understanding here and you're acting more like a teacher rather someone to answer my question which I really appreciate by the way. So its good to think of surrounding the angle with a triangle and then finding out the tangent angle of the triangle?
arildno
#12
Jul29-10, 01:25 PM
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No, let us forget about that angle for now. Okay?

1. The most important to remember about that slope, is that it tells you how steep the secant line is between the two points on the graph. Agreed?

2. Now, a tangent line at some point on the graph is that line which a) intersects with graph there, and b) is equally steep as the curve itself there.

3. When you perform the limiting operation (letting h be smaller and smaller) in order to calculate the derivative, you are simply calculating the slopes of successive secant lines and, in the end, when your two points overlap (h=0), sit back with the slope of the tangent line at (x,f(x)), i.e, the slope of the curve itself

4. So, the derivative at some point tells you how steep the curve is at that point.
zeromodz
#13
Jul29-10, 01:30 PM
P: 239
Quote Quote by arildno View Post
No, let us forget about that angle for now. Okay?

1. The most important to remember about that slope, is that it tells you how steep the secant line is between the two points on the graph. Agreed?

2. Now, a tangent line at some point on the graph is that line which a) intersects with graph there, and b) is equally steep as the curve itself there.

3. When you perform the limiting operation (letting h be smaller and smaller) in order to calculate the derivative, you are simply calculating the slopes of successive secant lines and, in the end, when your two points overlap (h=0), sit back with the slope of the tangent line at (x,f(x)), i.e, the slope of the curve itself

4. So, the derivative at some point tells you how steep the curve is at that point.
Okay thank you so much. I am going to take this info and start reading more about it. If I have any questions I will reply to this or PM you. Thank you so much.


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