How Do You Find the Equation of a Curve with a Given Gradient and Point?

[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.

 

[Hootenanny]

Well you know that;

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

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

~H

 

[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?

 

[Hootenanny]

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;<br /> <br /> \frac{dy}{dx} = 2x<br /> <br /> y = \int \frac{dy}{dx} dx = \int 2x dx<br /> <br /> y = x^2 + C<br /> <br /> Therefore, the equation of the curve is y=x^2 + C. In your case you are given a point, which will allow you to find the constant of integration.<br /> <br /> Do you follow?<br /> ~H

 

[dan greig]

so if i differentiate i get,

6x^2-9x-6

and then integrate,

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

 

[Hootenanny]

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

 

[dan greig]

so i get,

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

 

[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.

 

[Hootenanny]

so i get,

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

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

 

[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 ?

 

[Hootenanny]

Yes, that's 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

 

[dan greig]

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!

 

[Hootenanny]

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

Yes, I've corrected my post above.

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!

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

 

[VietDao29]

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.

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? :)

 

[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

 

[Hootenanny]

y = 2x^2 - 5x^2/2 - 6x + 3 ??

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

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

Yes, you are correct.

~H

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

 

[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

 

[dan greig]

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

 

[Hootenanny]

Yeah, you've got it. No problem.

~H