math help

letsbefree

Canada123 Posts

September 20 2008 04:36 GMT

Hi guys:

I have a math problem i need to solve, although it sounds easy but i just can't get the answer, can someone help me out?

Find the equation of the tangent line of y = x+3 (whole thing squared rooted) at point (6,3).

Please don't use derivatives to solve it. Thank you.=)

Nitrogen

United States5345 Posts

September 20 2008 04:41 GMT

just find some x values incredibly close to x=6, like x=5.999 and find the slope between that point and (6,3)

edit: oops i meant x values

paper

13196 Posts

September 20 2008 05:05 GMT

1/2 * ((6)+3)^(-1/2)

1/6?

i fail at math

evanthebouncy!

United States12796 Posts

September 20 2008 05:11 GMT

Oh that thing... Hang on I'll save you!

evanthebouncy!

United States12796 Posts

September 20 2008 05:12 GMT

http://www.math.hmc.edu/calculus/tutorials/limit_definition/

Read it, understand it, follow the example WITH A PICTURE LIKE YOU SHOULD DRAW A PICTURE(graph) AND SEE IF IT MAKES SENSE

If that still doesn't help shoot me a pm and we'll figure something out.

inkblot

United States1250 Posts

September 20 2008 05:18 GMT

The slope is 1/6, you can estimate this (since you're not using calculus) by doing what Nitrogen said, ie ( 3 - [ (5.99+3)^(1/2) ] ) / (.01) = .1667.....

Because (6,3) is 6 units in the x direction away from x = 0 and the slope is 1/6, the line will intersect the y axis one unit below (6,3), so it crosses at (0,2). Equation of line is y = (1/6)x + 2

Wonders

Australia753 Posts

September 20 2008 05:21 GMT

The only way I can think of of doing it without derivatives is something like:

1. Write the general equation for a line passing through (6,3), with the gradient as an unknown.
2. Find the points of intersection of that line and the y=sqrt(x-3) graph (the x values). Just write down the equation that x has to satisfy, it'll probably be a quadratic in x.
3. Find the values of the gradient so that the determinant of the quadratic is zero (so that the line only intersects the graph at 1 point). You'll probably get a few different values, so just look at what the line looks like for each value and pick the one that looks the best.

No idea why you'd do that, which would probably take a page or two of boring computations, when by taking the derivative you'd be able to do it in 2 lines though.

evanthebouncy!

United States12796 Posts

September 20 2008 07:33 GMT

Anyways take my advise as I'm a math tutor for my college and I just had alot of Math1A people asking me this question.

Nytefish

United Kingdom4282 Posts

September 20 2008 09:01 GMT

Well he did so no derivatives, so...

HeavOnEarth

United States7087 Posts

September 20 2008 11:16 GMT

#10

without derivatives you can only estimate where it's going , but it's really obvious after a while <_<
are you trying to just grasp the logic behind it?
if you understand it its essentially pointless to do it without the derivative
i mean, all u can do is guess and check numbers with the derivative right?

HeavOnEarth

United States7087 Posts

September 20 2008 11:24 GMT

#11

here's some extra practice if u want it
i'm taking calc BC senior year ; so we're in the same boat
http://archives.math.utk.edu/visual.calculus/2/definition.7/index.html

WITH A PICTURE LIKE YOU SHOULD DRAW A PICTURE(graph) AND SEE IF IT MAKES SENSE

definately, google some graph paper if u don't have any and plot it

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