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

• chwala
In summary: 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##
chwala
Gold Member
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.

Last edited:
chwala said:
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.

fresh_42 said:
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?

chwala said:
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.

chwala

## 1. What is the equation of the curve given the derivative dy/dx?

The equation of the curve can be found by integrating the given derivative with respect to x. This will result in the equation y = (kx-1)^6 + C, where C is a constant of integration.

## 2. How do you solve for k in the given equation?

To solve for k, you can use the given derivative to set up a system of equations. Substitute the x and y values of a point on the curve into both the derivative and the equation. This will result in an equation with one unknown variable, k, which can then be solved for.

## 3. What does the value of k represent in the equation?

The value of k represents the slope of the tangent line to the curve at any given point. It can also be thought of as the rate of change of the curve.

## 4. Can the given equation be used to find the area under the curve?

Yes, the given equation can be used to find the area under the curve by integrating the equation with respect to x over a given interval. This will give the definite integral of the curve, which represents the area under the curve within that interval.

## 5. Is it possible to find the x-intercepts of the curve using the given equation?

Yes, it is possible to find the x-intercepts of the curve by setting y = 0 in the equation and solving for x. This will result in the x-values where the curve intersects the x-axis.

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