# Equation of a curve homework

1. Apr 24, 2006

### 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.

2. Apr 24, 2006

### Hootenanny

Staff Emeritus
Well you know that;

$$\frac{dy}{dx} = (2x-3)(3x+2)$$

How do you suppose you can find $y=...$?

~H

3. Apr 24, 2006

### dan greig

i think i have to integrate the gradient but not sure how to.

how does this help me find the equation of the curve?

4. Apr 24, 2006

### Hootenanny

Staff Emeritus
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 $2x[/tex] then; $$\frac{dy}{dx} = 2x$$ $$y = \int \frac{dy}{dx} dx = \int 2x dx$$ $$y = x^2 + C$$ Therefore, the equation of the curve is [itex]y=x^2 + C$. In your case you are given a point, which will allow you to find the constant of integration.

Do you follow?
~H

5. Apr 24, 2006

### dan greig

so if i differentiate i get,

6x^2-9x-6

and then integrate,

2x^3-9x^2/2-6x ?

6. Apr 24, 2006

### Hootenanny

Staff Emeritus
You don't need to differentiate, you are given the gradient of the curve ($\frac{dy}{dx}$), so all you need to do is integrate.

~H

7. Apr 24, 2006

### dan greig

so i get,

(x^2 - 3x)(3x^2/2 +2x) ?

8. Apr 24, 2006

### tangents

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.

9. Apr 25, 2006

### Hootenanny

Staff Emeritus
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

10. May 1, 2006

### dan greig

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 ?

11. May 1, 2006

### Hootenanny

Staff Emeritus
Yes, thats correct. It may have been simpler to collect the terms before integrating thus;

$$6x^2 + 4x - 9x - 6 \equiv 6x^2 - 5x -6$$

But your working is correct, as long as you collect you terms before presenting you final answer that is fine.

~H

Last edited: May 1, 2006
12. May 1, 2006

### dan greig

do you mean 6x^2 - 5x -6

the result of this using x = 2 gives 8
does this mean,

-3 - c = 8

-c = 11,

c = -11 ????

sorry to keep on but i need to get this correct!

13. May 1, 2006

### Hootenanny

Staff Emeritus
Yes, I've corrected my post above.

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

14. May 1, 2006

### VietDao29

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.
---------------
Example:
Find the curve whose derivative is 4x3 + 5, and passes through the point (0, 1).
---------------
First, you integrate the derivative:
$$\int (4x ^ 3 + 5) dx = x ^ 4 + 5x + C$$
For every value of C, you'll have a function, and those functions that have the form x4 + 5x + C will have the derivative of 4x3 + 5 (they are parallel to each other, pick 2 distinct C's, graph it, and see what I mean, e.g x4 + 5x, and x4 + 5x + 4).
And 1 and only 1 of them will pass through (0, 1)
So:
1 = 04 + 5.0 + C
<=> C = 1.
So the curve is y = x4 + 5x + 1.
Can you get this? :)

15. May 1, 2006

### dan greig

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

16. May 1, 2006

### Hootenanny

Staff Emeritus
Almost, I think it is just a typo, but the first term should be 2x3.

Yes, you are correct.

~H

P.s. I edit my last post because the 2x2through me a bit.

17. May 1, 2006

### dan greig

sorry it was supposed to be

2x^3-5x^2/2-6x+c

so that gives me,

c = -3-16+10+12 = 3

18. May 1, 2006

### dan greig

sorry that was a bit of a marathon, thank you for your help

19. May 1, 2006

### Hootenanny

Staff Emeritus
Yeah, you've got it. No problem.

~H