I have a simple problem from calculus topics. Suppose we have
$x=at^2,\qquad y=2at$ and want to find $\dfrac{d^2y}{dx^2}$. There is given sample http://www.mathopolis.com/questions/a.php?id=137&ansno=957 I think that the answer is zero but here it is equal D why?
derivative of function
5 Answers
You are not taking the derivative of $y$ with respect to $t$, you are taking the derivative of $y$ with respect to $x$.
Now, by the Chain Rule we have that $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt},$ or, equivalently (solving for $\frac{dy}{dx}$), $\frac{dy}{dx} = \frac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}.$ Therefore, since $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$, and $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$, then $\frac{dy}{dx} = \frac{2a}{2at} = t^{-1}.$
Now you can repeat: from the Chain Rule we have $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \Biggl(\frac{d}{dx}\left(\frac{dy}{dx}\right)\Biggr)\frac{dx}{dt} = \frac{d^2y}{dx^2}\frac{dx}{dt}.$ Solving for $\frac{d^2y}{dx^2}$, we get $\frac{d^2y}{dx^2} = \frac{\quad\frac{d}{dt}\left(\frac{dy}{dx}\right)\quad}{\frac{dx}{dt}};$ since $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(t^{-1}) = -t^{-2}$, we have: $\frac{d^2y}{dx^2} = \frac{-t^{-2}}{2at} = -\frac{1}{2at^3},$ which is answer D in the given link.
Observe that $y^2 = 4ax$
Then $2y\frac{dy}{dx} = 4a$
So $\frac{dy}{dx} = \frac{2a}{y}$ [A]
Then $\frac{\mathrm d^2 y}{\mathrm dx^2} = \frac{-2a}{y^2} \frac{dy}{dx} =$ (using [A]) $-\frac{4a^2}{y^3} = -\frac{4a^2}{8a^3t^3} = -\frac{1}{2at^3}$
A slightly different approach not using the chain rule explicitly:
From $x=at^2$ you have $t=a^{-1/2}x^{1/2}$ so with $y=2at$ you have $y =2 a^{1/2}x^{1/2}$ so taking the derivative with respect to $x$ $\frac{\mathrm dy}{\mathrm dx}=a^{1/2}x^{-1/2}$ and doing it again and reusing $x=at^2$ $\frac{\mathrm d^2 y}{\mathrm dx^2}=-\frac{1}{2} a^{1/2}x^{-3/2}=-\frac{1}{2a t^3}. $
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0@Shai: True. Note that if $a \lt 0$ then $x \le 0$ so to deal with the point you would get $t=\sqrt{x/a}$, $y=2a\sqrt{x/a}$, $\mathrm dy / \mathrm dx =a\sqrt{1/ax}$ and $\mathrm d^2 y /\mathrm dx^2 =-\frac{1}{2} a\sqrt{1/ax^3} =-\frac{1}{2a t^3}$. – 2011-06-29
Here, you use the fact that
$\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$
Since you're doing second derivatives, you need a further differentiation:
$\frac{\mathrm d^2 y}{\mathrm dx^2}=\left(\frac{\mathrm dx}{\mathrm dt}\right)^{-1}\frac{\mathrm d}{\mathrm dt}\frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$
and this is the formula you should be using (replace $x$ and $y$ with the appropriate expressions).
I'm not sure what else to add to the explanation provided in your link. I think the key to understanding this question is the statement $\frac{dy}{dx}=\frac{\quad\tfrac{dy}{dt}\quad}{\tfrac{dx}{dt}},$ which follows from the chain rule: $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}.$ Do you understand why this is true?
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0That's a valid manipulation, but when you compute $\frac{dy}{dx}=\frac{d}{dx}\left(2at\right)=\frac{d}{dx}\left(2\cdot \frac{x}{t^2} \cdot t\right)=\frac{d}{dx}\left(\frac{2x}{t}\right)$ you now have to treat $t$ as a function of $x$, **not** as a constant. – 2011-06-29