Find the equation of the curve given: ##\frac{dy}{dx}=2(kx-1)^5##

[chwala]

Homework Statement

See attached textbook question and solution

Relevant Equations

Integration - Calculus

This is the question...hmmmm it stressed me a little bit.

Find the textbook solution here; no. 6

Now my approach to this was as follows;

On integration,

$y=\dfrac{(kx-1)^6}{3k} +c$

on using the point $(0,1)$ and $(1,8)$, we end up with

$1=\dfrac{1}{3k} +c$
$8=\dfrac{(k-1)^6}{3k} +c$

it follows that,

$8=\dfrac{(k-1)^6}{3k} + \dfrac{3k-1}{3k}$
...
$(k-1)^6+3k-1-24k=0$

$(k-1)^6-21k-1=0$

$k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1=0$

I let $P(k)=k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1$

using trial and error method from $±0, ±1...$ i got $P(3)=0$ implying $k=3$ is a factor,

thus substituting on;

$1=\dfrac{1}{3k} +c$

we get;

$1=\dfrac{1}{3×3} +c$

$1=\dfrac{1}{9} +c$

$c=\dfrac{8}{9}$

Therefore;

$y=\dfrac{(3x-1)^6}{3×3} +\dfrac{8}{9}$

$y=\dfrac{(3x-1)^6+8}{9}$

I would be interested if there would be another approach to this question. Cheers guys.

 

[fresh_42]

Homework Statement:: See attached textbook question and solution
Relevant Equations:: Integration - Calculus

I would be interested if there would be another approach to this question. Cheers guys.

With $t:=kx-1$ and integration you get quickly to $y=\dfrac{2}{6k}(kx-1)^6+c$ with two equations and two unknowns.

 

[chwala]

With $t:=kx-1$ and integration you get quickly to $y=\dfrac{2}{6k}(kx-1)^6+c$ with two equations and two unknowns.

Hi @fresh_42 ...that is the same approach that i used. Maybe i ought to be specific...is there a different way other than using the binomial expansion?

 

[fresh_42]

Hi @fresh_42 ...that is the same approach that i used. Maybe i ought to be specific...is there a different way other than using the binomial expansion?

Whatever we do, we will have the equation $21k=(k-1)^6-1.$ I did another substitution, $s:=k-1,$ got $s^6-21s-22=0$ where you can read of $s=-1$ immediately. I asked WA for the solution, but you can of course do the long division $(s^6-21s-22):(s+1)=-22 + s - s^2 + s^3 - s^4 + s^5$ and see whether you can guess $s=2,$ or check $\pm1, \pm 2, \pm 11$ as divisors of $22$ and hope that the solution is an integer solution.


At least it does not use binomial expansion, only division. But you cannot make the polynomial go away.