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Today I was doing some practice problems for the AP Calculus BC exam and came across a question I was unable to solve.

In the xy-plane, a particle moves along the parabola $y=x^2-x$ with a constant speed of $2\sqrt{10}$ units per second. If $\frac{dx}{dt}>0$, what is the value of $\frac{dy}{dt}$ when the particle is at the point $(2,2)$?

I first tried to write the parabola as a parametric equation such that $x(t)=at$ and $y(t)=(at)^2-at$ and then find a value for $a$ such that $\displaystyle\int_0^1\sqrt{(x'(t))^2+(y'(t))^2}dt=2\sqrt{10}$. However, since it was a multiple choice question we were probably not supposed to spend more than 3min on the question so I though that my approach was probably incorrect. The only information that I know for sure is that since $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\rightarrow\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$ and we are evaluating at $x=2$ and so $\frac{dy}{dt}=3\frac{dx}{dt}$. Other than that I am not sure how to proceed and so any help would be greatly appreciated!

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2 Answers 2

up vote 3 down vote accepted

You can't assume $x$ increases at a constant rate so saying $x=at$ isn't guaranteed to work.

Instead, observe the constant speed means that

$$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=2\sqrt{10}. \tag{*}$$

We also have by implicit differentiation,

$$\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$$

so that at $x=2$ we can say $\displaystyle \frac{dx}{dt}=\frac{1}{3}\frac{dy}{dt}$. Plug this into $(*)$ and see if you can solve for $\displaystyle\frac{dy}{dt}$...

(You also need to see if $y\,'$ is positive or negative independently.)

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Sorry if this seems like a dumb question, but why is it that a constant speed means $\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=2\sqrt{10}$. Note I used the same method as you give here to get a value which was the correct answer, but I thought it was wrong so refrained from mentioning it in the question. –  E.O. Apr 24 '12 at 8:12
    
@Emile: The speed of the curve's parametrization is the left-hand side (this is a generic formula), while the given quantity for the speed (which is constant) is the right-hand side. They're both the speed so of course they're equal! –  anon Apr 24 '12 at 8:24
    
Ahh.. I forgot... Is the left hand side simply the absolute value of the velocity vector which is the speed? –  E.O. Apr 24 '12 at 8:34
    
@Emile: The $\|\cdot\|_2$-norm, or magnitude, of the velocity vector, yes. –  anon Apr 24 '12 at 8:35

As you noticed, $dy=(2x-1)dx$. The speed is constant, so that $\dot{x}^2+\dot{y}^2=40$. So you get the system $$\begin{cases} \dot{y}=(2x-1)\dot{x} \\\ \dot{x}^2+\dot{y}^2=40 \end{cases}.$$ By substitution, $\left[1+(2x-1)^2 \right]\dot{x}^2 = 40$, whence, at $x=2$, $\dot{x}^2=4$. Since $\dot{x}>0$, you find $\dot{x}=2$, and then $\dot{y}=(2x-1)\dot{x}=3 \cdot 2 = 6$. I hope I did not make mistakes.

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Hm.. that was strange. I did the exact same method and got the same answer, but I dismissed it as being wrong because the logic I used to get the assumption that $(x'(t))^2+(y'(t))^2=40$ seemed strange. –  E.O. Apr 24 '12 at 8:07
    
No, it isn't. Speed (or better velocity) is a vector in the euclidean space, the derivative in time of the law of motion. Being constant, this vector has a fixed length, which is $\sqrt{\dot{x}^2+\dot{y}^2}$. –  Siminore Apr 24 '12 at 9:36

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