Derivatives

audiobliss

I'm trying to write a short paper (several paragraphs at most) on the history of the derivative for math class. I'm thinking about briefly discussing several of the mathematicians who contributed to the development of the derivative. Anyhoo, I'm having a hard time finding anything about it online. Do any of y'all know of a good resource online I could check out?

Thanks!

rskarvan

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/The_rise_of_calculus.html

bobman1235

The Wikipedia page goes over some of the history as well.

simm

Audio- I have a copy of the VNR Concise Encyclopedia of Mathematics you can have if you want. We are moving Tuesday to Cary so if you want it you need to get today or tomorrow. I live on Yeaton Glen Dr off Glenn Hi Road. Not sure how far that is from you but Kville is not that big. PM me if interested and I can give you my number and directions.

audiobliss

Thanks for all the links, guys. I think I've got it under control now!

simm - Thanks for the offer. You have a PM!

audiobliss

On a side note...

I know how you find a tangent to a curve at a point. However, how do you find a tangent to a curve at a point that also goes through a specified not on that curve? For instance, if you have the function y=x^2 and the point (1,1/2), how would you go about finding the line that's tangent to the function and goes through the specified point?

Thanks!

gatemplin

audiobliss wrote:

On a side note...

I know how you find a tangent to a curve at a point. However, how do you find a tangent to a curve at a point that also goes through a specified not on that curve? For instance, if you have the function y=x^2 and the point (1,1/2), how would you go about finding the line that's tangent to the function and goes through the specified point?

Thanks!

In that case, there are two lines tangent to the curve that pass through (1,1/2)

You have two points given: (1,1/2) and (x,x^2)

Find the slope using these points, and the derivative of y=x^2.
Equate them, solve for both of the x values.
Plug both x values back into y'=2x to solve for the slope of each line.
Use those points, to find the two tangent lines.

audiobliss

Let me see if I understand this correctly.

I use the point-slope formula and plug in the two points using the derivative of y=x^2 as the slope: x^2- 1/2 = 2x(x-1).

Then I solve for x which gives me x=-1/2 and x=3/2.

That means that the two points on my curve are (-1/2, 1/4) and (3/2, 9/4).

That means the equations for the two lines are y=-x - 1/4, and y=3x - 9/4.

Is that correct?

EDIT - Nope, I know what I did is not right. I'm trying to find what I did wrong.

audiobliss

Alright, here's what I've got so far. The parabola has to go through the point (-100, 100), so I assume that the equation for the function is y=(x^2)/100. Is that right?

That means the derivative is y'=x/50.

So, when I plug those into the point-slope formula I get the points x=-50 and x=52/100.

What am I doing wrong? It's quite obvious from looking at a graph that a line from either of those points to (100, 50) is going to be a secant line, not a tangent line.

gatemplin

That eqn for the parabola doesn't seem right, does it open up of down, is the vertex at (-100,100)?

Derive the equation using y=ax^2+bx+c

audiobliss

y=(x^2)/100 opens upwards and its origin is at (0,0). The point (-100, 100) has to be on the parabola, and it is on the curve y=(x^2)/100.

For this particular problem I'm only concerned with the tangent as you follow the curve from the left to the right. In other words, I only want to find the tangent with the positive slope.

audiobliss

Ok. I'm really not seeing what I'm doing wrong here.

If the parabola has to go through the point (-100, 100), then the only function it could be is y=(x^2)/100. From that y'=x/50. The tangent to this curve has to go through the point (100, 50), and it has to have a positive slope. So, x/50 is the slope I'm going to use. The points I'm going to use are (100, 50) and (x, (x^2)/100). Now, if I use the point-slope formula:

y(2)-y(1)=m(x(2)-x(1))

then plug in:
(x^2)/100 - 50 = (x/50)(x - 100)

then distribute:
(x^2)/100 - 50 = (x^2/50) - 100x/50

then simplify:
(x^2)/100 - 50 = (x^2)/50 - 2x

then group like terms together:
(x^2)/50 - 2x - (x^2)/100 = -50

then simplify:
2(x^2)/100 - (x^2)/100 -2x = -50

then combine like terms:
(x^2)/100 - 2x = -50

then solve for x:
x(x/100 - 2) = -50
x= -50 or x = (-50 + 2)100 = -4800

What'd I do wrong?

bobman1235

audiobliss wrote:

Ok. I'm really not seeing what I'm doing wrong here.

If the parabola has to go through the point (-100, 100), then the only function it could be is y=(x^2)/100.

I honestly would have to spend hours studying to remember the REST of your problem, but that statement is definitely not true. One that comes to mind is
y = (x+100)^2 + 100.

audiobliss

Haha, you're definitely right. My bad, there. But, this parabola has to go through the origin, so that kinda limits it.

gatemplin

Bobman might be right, I'm a little rusty on the parabolas. If all you are given is one point, you cant find the equation of a parabola. You may have missed some info.

audiobliss wrote:

What'd I do wrong?

I solved for x quickly and got ~29.3 and ~170.7

audiobliss

I can't even make it come out right when I use this same method on a simple problem.

If I just go with y=x^2 and the point (1, 1/2), then this is what I'd have:

1/2 - x^2 = 2x(1 - x)

When I work that out, I come up with x = 1/2 or x = 3/2. Neither of the tangents to y=x^2 at those two points even include the point (1, 1/2). I'm totally at a loss here, 'cause it seems to me that the method should work, but I can't make it work.

audiobliss

*sigh*

How did you solve it quickly is what I want to know!

gatemplin

Quadratic formula.

audiobliss

Ok. Now I feel really stupid (and maybe I should), but how did you use the quadratic formula in this problem?

And I still can't figure out why I can't get the right answer!! *shoots himself in the foot and calls in sick*

gatemplin

To solve for x from (x^2)/100 - 2x = -50

That gives you the x-coordinate of the two tangent points.

audiobliss

Why can't I just factor out the x and have x(x/100 -2) = -50? What's wrong with that?

gatemplin

You cant ignore just one of the x variables like that.

You have an equation 0=x^2-200x+5000 now plug that into the quad formula. Plug those x-values back into 0=x^2-200x+5000 and you should get 0=0

audiobliss

And then I can plug those x-vaules back into my original equation and come up with the coordinates for the points where the tangents are supposedly tangent to the curve, right?

gatemplin

Yeppers. And sub those x-values into y' to solve for the slopes of the respective tangent lines. Then use the points and slopes to solve for y intercepts and you have the line equations.

audiobliss

I think I wasn't plugging the x-values into y' to find the slope of the tangent. I'm pretty sure I've got it now.

Thanks a million!!!!!!

madmax

Geeks...
madmax

gatemplin

AB,

A really good book for understanding Calculus better, especially the history is A Tour of the Calculus by David Berlinski.

audiobliss

Thanks for the recommendation, gatemplin. And thanks for your help with the problem! xinfinity!

gatemplin

audiobliss wrote:

Thanks for the recommendation, gatemplin. And thanks for your help with the problem! xinfinity!

No problem, anytime.

Graham