A point moves along a parabola y^2 = 3x. Find the Approximate change in its distance from the origin as its x coordinate changes from 1 to 1.1

The answer is 0.125

I assume origin is (0,0) ( I am not sure if this is correct; maybe it meant origin of the original?) Anyway so I try:

At point 1:

y^2 = 3(1); I get the coordinates (1, sqrt(10) ) with a distance of sqrt(17) from the origin)

At point 2:

y^2 = 3(1.1); I get the coordinates (1.1, sqrt(3.3) ) with a distance of sqrt(2.12376) from the origin)

The difference between both distances are 1.0386...

Why am I getting wrong answer? Why "Approximate"?

  • 1
    $\begingroup$ You are correct that the origin is $(0,0)$. However, point number $1$ is at $(1, \sqrt{3})$, not $(1, 4)$. $\endgroup$
    – Arthur
    Jun 14, 2015 at 10:15
  • $\begingroup$ right! changed stuff. but still not getting right answer :( $\endgroup$
    – james
    Jun 14, 2015 at 10:23

1 Answer 1


Distance from origin is $\sqrt{x^2+y^2} = \sqrt{x^2+3 x}$. Let $x_0=1$, $h=0.1$; then the change in distance from the origin is

$$\sqrt{(x_0+h)^2+3 (x_0+h)} - \sqrt{x_0^2+3 x_0} \approx h \left [\frac{d}{dx} \sqrt{x^2+3 x} \right ]_{x=x_0} = \frac{2 x_0+3}{2 \sqrt{x_0^2+3 x}}h = 0.125$$

Typically, problems asking for small changes in a function in response to small changes in their arguments are asking for the evaluation of a derivative of the function.

  • $\begingroup$ It would be a better estimate if you evaluated the derivative in-between, so at $x_{1/2} = 1.05$. A bit more difficult to calculate, though. $\endgroup$
    – Arthur
    Jun 14, 2015 at 10:32
  • $\begingroup$ @Arthur: correct that centered differences provide an $O(h^2)$ estimate. However, the OP was looking for a particular answer, and this is standard Calc I fare. $\endgroup$
    – Ron Gordon
    Jun 14, 2015 at 10:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .