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equation of a curve |
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Apr24-06, 05:58 PM
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#1
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dan greig is
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equation of a curve
i really need some help with this question, my assignment is due in tommorow, i don't really understand the question - please help!
A curve has a gradient (2x-3)(3x+2) at the point (x,y) and passes through the point (2,-3). Find the equation of the curve. illustrate the answer with a sketch.
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Apr24-06, 06:07 PM
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#2
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Hootenanny is
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Well you know that;
How do you suppose you can find  ?
~H
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Apr24-06, 06:18 PM
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#3
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dan greig is
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i think i have to integrate the gradient but not sure how to.
how does this help me find the equation of the curve?
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Apr24-06, 06:23 PM
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#4
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Hootenanny is
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When you differentiate a curve, you obtain the gradient. Therefore, if you integrate the gradient you will obtain the curve.
Here's a simple example.
If the gradient of a curve is given as  then;
Therefore, the equation of the curve is  . In your case you are given a point, which will allow you to find the constant of integration.
Do you follow?
~H
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Apr24-06, 06:31 PM
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#5
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dan greig is
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so if i differentiate i get,
6x^2-9x-6
and then integrate,
2x^3-9x^2/2-6x ?
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Apr24-06, 06:32 PM
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#6
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Hootenanny is
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You don't need to differentiate, you are given the gradient of the curve (  ), so all you need to do is integrate.
~H
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Apr24-06, 06:41 PM
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#7
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dan greig is
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so i get,
(x^2 - 3x)(3x^2/2 +2x) ?
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Apr24-06, 07:08 PM
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#8
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tangents is
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If you distribute before you integrte you get 6X^2-5x-6
if you integrate that you get 2x^3-(5/2)x^2-6x+c
Plug in you x value to find c.
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Apr25-06, 12:04 PM
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#9
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Hootenanny is
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Originally Posted by dan greig
so i get,
(x^2 - 3x)(3x^2/2 +2x) ?
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You cannot integrate the brackets seperately as you have done. I would recommend expanding the brackets as tangents suggests, then integrating. As I said before, once you have integrated, you can use the given point to find the constant of integration.
~H
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May1-06, 11:11 AM
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#10
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dan greig is
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if i expand the brackets from (2x-3)(3x+2) i get,
6x^2 + 4x - 9x - 6
and then integrate,
y = 2x^3 + 2x^2 - 9x^2/2 - 6x +c
do i then just plug in the x and y values to find c by rearranging,
y - c = 2x^3 + 2x^2 - 9x^2/2 - 6x ?
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May1-06, 11:35 AM
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Last edited by Hootenanny; May1-06 at 11:51 AM..
#11
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Hootenanny is
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Yes, thats correct. It may have been simpler to collect the terms before integrating thus;
But your working is correct, as long as you collect you terms before presenting you final answer that is fine.
~H
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May1-06, 11:44 AM
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#12
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dan greig is
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do you mean 6x^2 - 5x -6
the result of this using x = 2 gives 8
does this mean,
-3 - c = 8
adding 3,
-c = 11,
c = -11 ????
sorry to keep on but i need to get this correct!
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May1-06, 11:55 AM
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#13
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Hootenanny is
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Originally Posted by dan greig
do you mean 6x^2 - 5x -6
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Yes, I've corrected my post above.
Originally Posted by dan greig
the result of this using x = 2 gives 8
does this mean,
-3 - c = 8
adding 3,
-c = 11,
c = -11 ????
sorry to keep on but i need to get this correct!
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No, you have to plug you numbers (x = 2, y = -3) into your intergrated equation. In my previous post I was simply pointing out that it would have been easier to integrate the equation if you had simplified first.
~H
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May1-06, 12:26 PM
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#14
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VietDao29 is
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Homework Helper
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Originally Posted by dan greig
i really need some help with this question, my assignment is due in tommorow, i don't really understand the question - please help!
A curve has a gradient (2x-3)(3x+2) at the point (x,y) and passes through the point (2,-3). Find the equation of the curve. illustrate the answer with a sketch.
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Okay, I think I'll make it a bit clearer for you.
A curve has a gradient (2x-3)(3x+2) at the point (x,y) simply means that the slope of the tangent line to the function at the point (x, y) is (2x-3)(3x+2). Or in other words, the derivative of that function is (2x-3)(3x+2). Can you get this?
So what you should do is to find the function by knowing its derivative, and 1 point it passes through.
You should first find the functons whose derivative is (2x-3)(3x+2), you can do this by integrating the derivative. And only 1 of those functions will pass the point given, with this information, you'll be able to obtain the desired function.
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Example:
Find the curve whose derivative is 4x 3 + 5, and passes through the point (0, 1).
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First, you integrate the derivative:

For every value of C, you'll have a function, and those functions that have the form x 4 + 5x + C will have the derivative of 4x 3 + 5 (they are parallel to each other, pick 2 distinct C's, graph it, and see what I mean, e.g x 4 + 5x, and x 4 + 5x + 4).
And 1 and only 1 of them will pass through (0, 1)
So:
1 = 0 4 + 5.0 + C
<=> C = 1.
So the curve is y = x 4 + 5x + 1.
Can you get this? :)
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May1-06, 03:24 PM
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#15
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dan greig is
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Posts: 41
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would my final equation be,
y = 2x^2 - 5x^2/2 - 6x + 3 ??
i also need to sketch the graph, to find the points where the line intercects the axis do i use,
y = 0 for x axis
x = 0 for y axis
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May1-06, 03:53 PM
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#16
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Hootenanny is
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Posts: 8,929
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Originally Posted by dan greig
y = 2x^2 - 5x^2/2 - 6x + 3 ??
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Almost, I think it is just a typo, but the first term should be 2x 3.
Originally Posted by dan greig
i also need to sketch the graph, to find the points where the line intercects the axis do i use,
y = 0 for x axis
x = 0 for y axis
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Yes, you are correct.
~H
P.s. I edit my last post because the 2x 2through me a bit.
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