# Derive or Integrate

Discussion in 'Math' started by KCHARROIS, Jan 30, 2014.

1. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Hello,

So I keep stumbling on the problem on when should I differentiate or integrate. If someone told me to integrate an equation I could do it but if he or she gave me the same question without telling me to integrate I would not know if I should integrate or differentiate? So how can you tell? For example, finding the voltage across a capacitor at specific time, I would derive right?

Thanks

2. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
When would you want to integrate an expression?
When would you want to differentiate an expression?

Integration means taking the sum. Graphically it is calculating the area under a curve.
If you want to determine how much ice cream you consume in an entire year you integrate your daily consumption of ice cream.

Derivative or differentiation means taking the difference. Graphically it means finding the slope of a curve at a given point. If you want to find how much your body weight is increasing or decreasing on a daily or monthly basis by consuming too much ice cream or by exercising, you take the derivative, that is the difference from day to day or month to month.

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3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
are you sure?

Let's try

Here is an equation

y = 2x + 3

How would you integrate it?

5x+20 = 0

Let us think about why you would integrate or differentiate. A good knowledge of this will clear up your problem.

Why do you integrate and why do you differentiate?

4. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Integrating

y = 2x + 3
= x^2 + 3x

5x+20 = 0
5/2x^2 + 20x = 0

I know that this isn't the correct way of writing it down but in short they're the answers.

You integrate the find the area under a curve and you differentiate to find the slope of a curve.

5. ### studiot AAC Fanatic!

Nov 9, 2007
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So the first equation says x = -4, does this satisfy the second?

My point is that whilst it is possible to integrate equations, this is not taught at elementary level.

Perhaps you need to start back a step and revisit what an equation is?

6. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Ok well y = 2x + 3 is a function while the other equation is just algebraic where x = 4.

7. ### studiot AAC Fanatic!

Nov 9, 2007
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I am not being patronising, but can you not see what I mean here?

If not I will have to find another way of putting it because you are just plain wrong, because you are suffering under a misunderstanding.

8. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I understand if your frustrated with me but that's what I mean if someone tells me to integrate I can integrate but if I were told that's either you integrate or differentiate I wouldn't know how to choose between the 2.

9. ### poopscoop Member

Dec 12, 2012
139
16
Fear not. No one is going to give you an equation and expect you to know what to do. Rather, they'll give you the equation for spring force and ask"How much work is required to move the spring from x=0 to x=5?"

Try it. F=kx and let k=5

Now try to find the velocity of an object who's position can be described by p=2x^2+2.

Your teacher is dropping the ball by not giving you word problems.

10. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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I am not frustrated with you. I see that your problem goes much deeper than integrate v differentiate (no offence meant).

You have not been able to successfully integrate either of the two simple equations I gave.

So let us go back to some definitions.

An expression is a combination of mathematical quantities and symbols.

So
${x^3}$

${x^{\frac{1}{5}}} + 3$

$\frac{{\sin x}}{{2x}}$

$\sqrt {{b^2} - 4ac}$

$x + y = 3$

$y = {x^2} - 15x + 3$

are all mathematical expressions.

An equation is an expression containing an equals sign.

So the last two above are equations.

An expression is algebraic if it only contains powers, roots and the four operations +,-,*,/

All the above are algebraic except the third which is partly trigonometric.

Do you understand this so far

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11. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Yes I understand this.

12. ### MrChips Moderator

Oct 2, 2009
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studiot is giving you an excellent lesson here.

Note that in my post I did not use the word "equation". I made a point of saying "expression".

studiot is making a point of making sure you understand the difference between an equation and an expression.

3x^2 + 5x + 4

would be an expression.

y = 3x^2 + 5x + 4

is an equation, y is the expression.

We don't integrate or differentiate an equation.

dy/dx is the derivative of the expression y.

∫y.dx is the integral of the expression y with respect to x.

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13. ### studiot AAC Fanatic!

Nov 9, 2007
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So try to keep the correct use of these terms in mind.
It really will help, that is why they were chosen in the first place.

Now you manipulate expressions according to certain rules.
Equations have additional rules and one in particular is very important.

Whatever you do to the left hand side (LHS) of an equation you do to the right hand side (RHS) to maintain the equality.

Expressions (that are not equations) do not have 'sides' or only have one side.

So in my examples 5x+20=0 is an equation.

If we were to integrate this we must integrate both sides separately

So what is the integral of 0?

similarly for y = 2x+3

What is the integral of y?

A good simplification to take forward at the moment is

We would normally expect to differentiate an equation and integrate an expression if calculus is called for.

The exception is that the equation must have a minimum of two variables so my first example does not qualify.

But you can differentiate the second to find

$\frac{{dy}}{{dx}} = 2$

and integrate the first two expressions in my list to find

The integral of ${x^3}$is ${x^4}+C$
and
the integral of

${x^{\frac{1}{5}}} + 3$

is

$\frac{{5{x^{\frac{1}{5}}}}}{6} + 3x + C$

Last edited: Jan 30, 2014
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14. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Wow things are now rolling in my head. I now understand why we use integration to find voltage across a capacitor with constant current.

I = V/R
V = Q/C

Therefor I = Q/RC

And from a known "equation" -I = dq/dt

We get -Q/RC = dq/dt

Variables on both sides means that I must integrate.

∫-Q/RC = ∫dq/dt

Am I correct?

15. ### studiot AAC Fanatic!

Nov 9, 2007
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Yeah! You are getting there!

There is more so don't run too quickly.

It will take me about 20 mins to prepare the next reply.

16. ### Sparky49 Well-Known Member

Jul 16, 2011
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417
Sorry to cut in here, but this has been useful to me. Thanks studiot and Mr Chips. Definitely an "ah, that's cool" moment.

17. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Things are so much more clearer now. And ok sounds great Studiot!

18. ### studiot AAC Fanatic!

Nov 9, 2007
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515
There are several ways to regard integration and differentiation.
Mr Chips has already offered one way.
You seem to have gathered another which is more useful in these circumstances.

Differentiation and integration can be regarded as the reverse of each other.

That is differentiation 'undoes' the operation of integration and integration 'undoes' the operation of differentiation.

Consider my first equation again.

5x+20=0
5x=-20
x=-4

We can 'solve' this directly.
By solve I mean find a value for x that we want.

Now consider Ohm's Law

V=IR

If we know the current through a given resistor we can directly calculate the voltage.

But

for a capacitor

V = 1/C∫Idt and I = dQ/dt

and for an inductor

V = LdI/dt

For a resistor, capacitor and inductor in series we form the equation

${V_t} = IR + L\frac{{dI}}{{dt}} + \frac{1}{C}\int\limits_0^t {Idt}$

Which gives the voltage at time t

Now this equation is not very useful as it stands, but if we note that $\int\limits_0^t {Idt}$ is the charge that has accumulated on the capacitor (assuming it was initially zero) by time t.

We do this by integrating the expression above (as you noted but between limits) so we can replace our integral by Q the charge on the capacitor and the equation becomes

${V_t} = IR + L\frac{{dI}}{{dt}} + \frac{Q}{C}$

Multiply through by C

$C{V_t} = IRC + LC\frac{{dI}}{{dt}} + Q$

But this is a mixture of variables, voltage, current and charge so differentiate with respect to time

$C\frac{{d{V_t}}}{{dt}} = RC\frac{{dI}}{{dt}} + LC\frac{{{d^2}I}}{{d{t^2}}} + \frac{{dQ}}{{dt}}$

Finally replace dQ/dt with I

$C\frac{{d{V_t}}}{{dt}} = RC\frac{{dI}}{{dt}} + LC\frac{{{d^2}I}}{{d{t^2}}} + I$

Which is a second order differential equation that can be solved if V is known as a function of t.
Solving a differential equation is another word for integrating it to transform the derivatives to explicit expressions we can calculate with.

This sequence shows where to differentiate and where to integrate.

The differential equation method of solution is not widely used for circuits, you are probably more familiar with other methods.

Last edited: Jan 30, 2014
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19. ### MrChips Moderator

Oct 2, 2009
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If you are now learning about differentiation and integration then ignore the rest of this post until the time comes when you are ready for this.

This is a perfect introduction to Laplace Transforms.

As you get into more advanced topics requiring you to solve RCL circuits as in filter circuits for example, you will be required to solve problems involving differential equations. These will involve both derivatives and integrals. Studiot has given you one such example above.

Here is a pdf doc that gives an excellent introduction to Laplace Transforms and how it is applied to analyzing RCL filter circuits.

A Laplace Transform Cookbook

Save this for future reference.

20. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Hi MrChips,

I have taken classes on derivatives and integration and a combination of both introducing laplace transforms which I do agree works great for RLC circuits but I'm definitely going to have to go through my text book restarting from the beginning. I have the book called Basic Technical Mathematics with Calculus, 9th edition.

Thanks