Help for quickest and clearest route for differentiation question

  • Thread starter Thread starter kauymatty
  • Start date Start date
  • Tags Tags
    Differentiation
AI Thread Summary
To find the coordinates of points on the curve y=1/3x^(3/2)-x^(1/2) where the gradient equals 3/4, the discussion focuses on solving the equation x^(1/2)-x^(-1/2)=6/4. A suggested method is to substitute z = x^(1/2) and transform the equation into a standard quadratic form. After eliminating square roots, the challenge remains with negative powers, and multiplying through by a common denominator like 16x is proposed as a solution. The conversation emphasizes algebraic manipulation to isolate x effectively.
kauymatty
Messages
2
Reaction score
0

Homework Statement


Find the coordinates of the points on these curves at which the gradient has the given values.

y=1/3x^(3/2)-x^(1/2), gradient=3/4


The Attempt at a Solution



Basically the problem I have isn't really finding the answer because...

x^(1/2)-x^-(1/2)=6/4[/b] and then from here I just used trial and error to find 4, but I want to know how I would follow it through with the algebra, i.e. only has x on one side as one term e.g. x= ans.


Thanks in advance,

kauymatty
 
Physics news on Phys.org
Try substituting z = x^(1/2), then cast the result into the form az^2 + bz + c = 0.
 
I'd square both sides of the equation. That'll eliminate the square roots completely.
 
vela said:
I'd square both sides of the equation. That'll eliminate the square roots completely.

Thanks vela and banders, but after eliminating the roots I would still have the negative power, would I just get a common denominator by multiplying through with 16x as the denominator? (big of a noobish question XD)
 
Multiplying through by 16x would be fine.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top