derivative of function

Question

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?

Answer 1

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.

Answer 2

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]) $-\frac{4a^2}{y^3} = -\frac{4a^2}{8a^3t^3} = -\frac{1}{2at^3}$

Answer 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 $x=at^2$ $$\frac{\mathrm d^2 y}{\mathrm dx^2}=-\frac{1}{2} a^{1/2}x^{-3/2}=-\frac{1}{2a t^3}. $$

Answer 4

In general, if one has the parametric equations $$ x=f(t),\quad y=g(t)\tag{1} $$ one can^$\dagger$ write $\frac{dy}{dx}$ in terms of $t$ by $$ \frac{dy}{dx}=\frac{g'(t)}{f'(t)}\tag{2}=:h(t)\;. $$

Now, to write $\frac{d^2y}{dx^2}$ in terms of $t$, one simply observes that $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})$ and replaces (1) by $$ x=f(t),\quad \frac{dy}{dx}=h(t)\tag{3} $$ to get $$ \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{h'(t)}{f'(t)}\tag{4} $$

The calculation is of course essentially the same as in Arturo's answer.

Sometimes, people use Newton's notation for derivatives in (2) and write $$ \frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}\quad \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)\frac{1}{\dot{x}} = \frac{\dot{x} \ddot{y}-\dot{y} \ddot{x}}{\dot{x}^{3}} $$ This is related to the notion of curvature.

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^{$\dagger$ Assume that $f$ and $g$ are both smooth.}

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So if $f(t)=at^2$ and $g(t)=2at$, combining (2) and (4) together, one has $$ h(t) = \frac{2a}{2at} =\frac1t,\quad \frac{h'(t)}{f'(t)} =-\frac{1}{t^2}\cdot\frac{1}{2at}=-\frac{1}{2at^3}\;. $$

Answer 5

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).

Answer 6

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?