Quod Erat Demonstrandum

2009/09/24

Differentiation of parametric equations

It is extremely easy to set up questions about differentiation techniques (but good real life application questions are really rare, esp. at secondary school level), apart from tedious computation, when the differentiation involves parameter, students may have difficulties, like mistaking:

\frac{d^2y}{dx^2} = 1/\frac{d^2x}{dy^2} (wrong!)
\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}/\frac{d^2x}{dt^2} (wrong!)

Here is a question in recent quiz, which involves parametric equations:

Let

x = \sqrt{2 + t^2}
y = \sqrt{2 - t^2}

Show that \frac{d^2y}{dx^2} = -\frac{4}{y^3}.

Quite a neat result, isn’t it?

Of course, if students could eliminate t and obtain x^2 + y^2 = 4 at the beginning, the solution path is easy.

But, many tried to find \frac{dy}{dt} and \frac{dx}{dt} and they gave

\frac{dy}{dx} = \frac{-t(2-t^2)^{-1/2}}{t(2+t^2)^{-1/2}}

then, when they tried to differentiate the above directly, many were at a loss with variables x, y and t.

Not many students could move one step further to see the following “bridging" step:

\frac{dy}{dx} = -\frac{x}{y}

Differentiate once more, yield \frac{d^2y}{dx^2} = -\frac{4}{y^3}.

How to set a question so as to obtain the neat result \frac{dy}{dx} = -\frac{x}{y}?

Well, we may use the “walking backward" strategy, start with

\frac{dy}{dx} = -\frac{x}{y}

We then solve the above differential equation

ydy = -xdx
\int ydy = -\int xdx
\frac{y^2}{2} = -\frac{x^2}{2} + C_1
x^2 + y^2 = C

It is just an equation of circle, and it is not that difficult to re-write the equation into parametric form, like

x = \sqrt{C + t^2}
y = \sqrt{C - t^2}

or even

x = r\cos\theta
y = r\sin\theta

etc.

A bit modification, we may consider an ellipse, say

4x^2 + 9y^2 = 1

and it is easy to have \frac{dy}{dx} = -\frac{4x}{9y}

Now, “walking backward", we can establish the parametric equation (one could choose as ugly as he likes) of the ellipse as

x = \sqrt{\frac{1}{8} + 9t^2}
y = \sqrt{\frac{1}{18} - 4t^2}

By the “set-in-advance" result

\frac{dy}{dx} = -\frac{4x}{9y}
4x + 9y\frac{dy}{dx} = 0

Differentiate with respect to x, yield

y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 = -\frac{4}{9}

Well, we may ask students to prove the above.

Urm, could we set up similar questions by considering the “set-in-advance" results, like

\frac{dy}{dx} = \frac{ax + by}{cx + dy}?

Students, try to explore it at your command.

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2 則迴響 »

  1. 我咁就冇左8分- –

    迴響 由 FTL — 2009/09/25 @ 1:36 上午 | 回應

  2. Hi John,

    Thanks for your meaningful article! Now students will be motivated to learn differential equations :)

    Cheers,
    Wen Shih

    迴響 由 Wen Shih — 2009/09/29 @ 5:41 下午 | 回應


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