Second derivative of parametric equation at given point.

Question

Let $f(t)=(t^2+2t,3t^4+4t^3), t>0$. Find the value of the second derivative, $\frac{d^2y}{dx^2}$at the point $(8,80)$

This is a past Math subject GRE question, and the usual formula: for second parametric derivatives $$\frac{\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}$$ took me much longer than 2.5 minutes (the average time per question) to compute. I'm thinking there has to be a faster way than actually computing all those partials, any ideas?

Thanks

Answer 1

Step 1 - Derivatives

Speed: Derivatives of polynomials in expanded form should be basically automatic for anyone doing/done an calculus course so the speed is basically as quickly as you write.

$\frac{dy}{dt}=12t^3+12t^2$ and $\frac{d^2y}{dt^2}=36t^2+24t$

$\frac{dx}{dt}=2t+2$ and $\frac{d^2x}{dt^2}=2$

Step 2: Find $t$

Speed: There is some work involved in finding $t$. However it only involves solving a quadratic so should be quick. The quadratic is not difficult so again basically as quickly as you can write it down.

Solving: $t^2+2t=8$ $\rightarrow$ $t^2+2t-8=0$ $\rightarrow$ $(t-2)(t+4)=0$ $\rightarrow$ $t=2$ as $t>0$.

Step 3 - Evaluate Expression (Edited slightly for no calculator comment)

Speed: Do not try and simply the algebraic expression for $\frac{d^2y}{dx^2}$. Just substitute in $t=2$. This is just arithmetic so should be pretty quick.

$$\left(\dfrac{d^2y}{dx^2}\right)_{t=2}=\left(\dfrac{\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}\right)_{t=2}=\dfrac{(36\cdot2^2+24\cdot2)\cdot(2\cdot2+2)-2\cdot(12\cdot2^3+12\cdot2^2)}{(2\cdot2+2)^3}$$

To calculator this efficiently in your head/paper use arithmetic factorizing to avoid having to deal with big numbers. Below is the steps I would be doing in my head/book (I certainly wouldn't be writing them all down).

$$=12\cdot\left(\frac{(3\cdot2^2+2\cdot2)\cdot6-2\cdot(2^3+2^2)}{6^3}\right)$$

$$=12\cdot2\cdot\left(\frac{(3\cdot2+2)\cdot6-(2^3+2^2)}{6^3}\right)$$

$$=12\cdot2\cdot\left(\frac{8\cdot6-12}{6^3}\right)$$

$$=12\cdot2\cdot6\cdot\left(\frac{8-2}{6^3}\right)$$

$$=\frac{6\cdot2\cdot2\cdot6\cdot6}{6^3}$$

$$=4$$

Personal Reflection

Having a BSc and interest in Mathematics I would expect myself to take about and 60 seconds for this question using the formula you provided. If I was to use @QuantumFool's guttural instinct technique it would be more like 3 minutes as the calculus/algebra becomes more involved.

Obviously different people will have different experiences.

Answer 2

Probably easier to say now than during an examination !

I suppose that the idea was to work the numerator

$$x=t^2+2 t\qquad , \qquad x'=2 t+2\qquad , \qquad x''=2$$ $$y=3 t^4+4 t^3\qquad , \qquad y'=12 t^3+12 t^2\qquad , \qquad y''=36 t^2+24 t$$ $$x' y''-y'x''=(2 t+2) \left(36 t^2+24 t\right)-2 \left(12 t^3+12 t^2\right)=48 t^3+96 t^2+48 t$$ $$x' y''-y'x''=48 t (t+1)^2$$ $$\frac{x' y''-y'x''}{(x')^3}=\frac{48 t (t+1)^2}{8(t+1)^ 3}=\frac{6 t}{t+1}$$

Now, since $t>0$, $$x=t^2+2t=8 \implies t^2+2t+1=9\implies (t+1)^2=9\implies t=2$$ we can just check that $t=2\implies y=80$. So, we have the point and the derivative is immediate.

Trying to immediately express $y$ as a function of $x$ is not much simpler $$x=t^2+2 t\implies x+1=(t+1)^2\implies t=\sqrt{x+1}-1\implies \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$

Answer 3

How I would think about this problem:

We want $(t^2+2t,3t^4+4t^3) = (8,80)$, so $t^2+2t = 8\implies (t+1)^2 = 9 \implies t+1 = \pm 3 \implies t = -1\pm 3 = 2$ or $-4$. So $t=2$.

Now $x=t^2+2t\implies \frac{dx}{dt} = 2t+2$ and $y = 3t^4+4t^3\implies\frac{dy}{dt} = 12t^3+12t^2$, so $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{12t^3+12t^2}{2t+2} = \frac{6t^2(2t+2)}{2t+2} = 6t^2.$$ Well that simplifies things a lot! We want $\frac{dy^2}{dx^2}$, and $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)/\left(\frac{dx}{dt}\right) = \left(\frac{d}{dt}(6t^2)\right)/(2t+2) = \frac{12t}{2t+2} = \frac{6t}{t+1}.$$ With $t=2$, this becomes $\frac{6(2)}{2+1} = \frac{6*2}{3} = \boxed{4}$.

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Comments: I don't like cumbersome formulas, so when asked to find $\frac{dy^2}{dx^2}$, my "instinct" would say to find $\frac{dy}{dx}$ first, and then take its derivative. I actually ran into this problem while practicing for the GRE, and when I first did it I made the mistake of taking the result $\frac{dy}{dx} = 6t^2$ and differentiating in $t$ to get the answer, essentially evaluating $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ instead of $\frac{d}{dx}\left(\frac{dy}{dx}\right)$. So in my biased view the trickiest part of the problem was to not make that mistake, since otherwise the problem seemed designed to make calculations not terrible (given that $\frac{dy}{dx}$ ended up being a nice, short expression).