Thread: Integral Calculus

1. find the equation of the curve which passes through the points (1,3), (2,9) and whose gradient is directly proportional to x((2x^2)-3)
the second question is;

2. given that the gradient of a curve is (ax-3) and the curve passes through the points (-1,8 ) , (3,4), find the equation of the curve.

thank you for your patience, and would appreciate any of your help

Is this homework? Even if it is, the only hit I'll give is that the function has to be some multiple of the inverse of the gradient.

nope, just doing some revision,
inverse the gradient? srry, but i dont really get that part, its supposed to be proportional but how come you inverse it?

Well if the gradient is the Partial derivative of a function with respect to x y z or whatever your variables are, then you need to integrate with respect to those variables to arive back at your function. Then there should be some multiple that will give you the function passing through those points.

This is just guessing, I had a problem like this once in one of my Calculus classes and I believe that is what I did.

Oo, yes but i tackled it in another way, by making it in a form of a=kx
after you integrate the equation, and your variable c, will have become 2c as you set the given gradient as a proportional value to your integrated equation, and substitute the co-ordinates given forming an algebraic equation and you solve for c
and later you get c as 2.5, and thus your c or arbitrary constant becomes 5, then you can insert the c=5 in to the equation, and therefore you get your equation of the original curve
im not sure if this is the actual way of solving it, but it sounds reasonable. your way i tried, it equally works though im abit confused

Originally Posted by Heinsbergrelatz

1. find the equation of the curve which passes through the points (1,3), (2,9) and whose gradient is directly proportional to x((2x^2)-3)
the second question is;

2. given that the gradient of a curve is (ax-3) and the curve passes through the points (-1,8 ) , (3,4), find the equation of the curve.

thank you for your patience, and would appreciate any of your help

Well x((2x^2)-3) = A(2x^3 -3x)+C, if that is the gradient then integrating it gives the curve.
So 2x^3 -3x whn intregrated gives A(8x^4 -6x^2) + C, as the curve
so y = A(8x^4 -6x^2) + C from the (1,3) we know when y=3 x=1
So 3= A(8 - 6) + C
or 3=2A+C

thus C=1 and thats the equation of the line.

For the (2,9) point then
9= A(16x8 -6x4 )=A(108-24) =84A+C

3=2A+C (i)
9=84A+C (ii)

from (i) C= 3-2A
put that in (ii)
9=84A + 3 - 2A

6=82A so

A=6/82

put that back in (i)

C=3- 12/82.=2+70/82

So is that the answer?

y = A(8x^4 -6x^2)

y=6/82(8x^4 -6x^2) + 2+70/82

Looks pretty horrible doesn't it?

I think maybe I should have said


y = A(8x^4 -6x^2 + C)

rather than

y = A(8x^4 -6x^2) + C

that might give a nicer answer.

make that maybe a definately

so
y = A(8x^4 -6x^2 + C)
goes through 1,3

3=A(2+C) (1)

and 2,9

9=A(84 +C)

from (1)

A=3/(C+2)

9=3(84+C)/(C+2)

well that looks worse

3= (84+C)/(C+2)

3C+6=84+C

2C=78

so C=39!!

And A=3/41!!


Giving y=3(8x^4 -6x^2+39)/41

Looks a bit better I suppose?

yes, i get where you get the idea and the point of the gradient function in this integral part but i have little better solution, correct me if its wrong thank you:

since its directly proportional,which is ;
m=gradient




as you see from the question two sets of points are given (1,3) (2,9)




if you derive these two sets of equations, yo get two variables (c,k),now you can perform the simultaneous linear equation here;



now you insert these two values or substitute them in to the integrated equation;



and there for you will now get the equation;

Removed due to my own stupidity

Why isn't the solution correct? The problem states exactly what he wrote down. that the gradient is directly proportional to the equation. Unless the gradient is a function of more variables than x then it seems that the solution would be to integrate. If it is a function of more than just x then I don't see how enough information is given to solve that problem.

ACK! by integral I meant derivative. it's been a bit since the last time I talked calculus. My bad, he got it. I'm off my rocker.