Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A particle is moving along the curve $y=3\sqrt{5x+6}$. As the particle passes through the point $(2,12)$, its $x$-coordinate increases at a rate of $3$ units per second. Find the rate of change of the distance from the particle to the origin at this instant.

I understand that $\dfrac{\mathrm dx}{\mathrm dt} = 3$, that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$, and that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy}{\mathrm dt} \cdot \dfrac{1}{3}$.

I also know that the derivative of $y = \sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$. I then get $\dfrac{dy}{dt} = \dfrac{15}{2\sqrt{5x+6}} \cdot \dfrac{dx}{dt}$. What steps do I take next? Did I make any errors in reasoning? Would 45/8 be correct as the $\dfrac{dy}{dt}$ term? I tried the suggestion in the answers, but it doesn't work. What do I do next? If someone provides a full step-by-step summary of how to solve, I might even catch my own mistake.

share|cite|improve this question
up vote 3 down vote accepted

If the question was asking about the rate of change in the distance from the particle to the $x$-axis, you'd be right on track. That's not what they want, though.

The distance from the particle to the origin is $$D=\sqrt{x^2+y^2}=\sqrt{x^2+\left(3\sqrt{5x+6}\right)^2}=\sqrt{x^2+9(5x+6)}=\sqrt{x^2+45x+54}.$$ The question wants $\frac{dD}{dt}$, i.e. $$\frac{dD}{dx}\cdot\frac{dx}{dt},$$ when $x=2$. Can you take it from there?

share|cite|improve this answer
I tried. Maybe I made a simplification error. I multiplied my answer by 3, the value of dx/dt, but it is still incorrect. (at least WebWorK marks it as incorrect). – cuabanana Mar 29 '13 at 19:19
We should get $$\frac{dD}{dx}=\frac{2x+45}{2\sqrt{x^2+45x+54}},$$ and so when $x=2,$ we have $$\frac{dD}{dt}=\frac{dD}{dx}\cdot\frac{dx}{dt}=\frac{49}{4\sqrt{37}}\cdot 3=\frac{147\sqrt{37}}{148}.$$ – Cameron Buie Mar 29 '13 at 20:14
Thank's I tried it . It's the correct answer. – cuabanana Mar 30 '13 at 23:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.