Quod Erat Demonstrandum

2019/05/05

What’s wrong?

Filed under: Additional / Applied Mathematics,NSS — johnmayhk @ 7:05 下午
Tags: ,

Here is a basic level M2 question:

Given that \sqrt{xy}=7+2y, find \frac{dy}{dx} at (-\frac{1}{3},-3).

Student 1 gave

\frac{1}{2\sqrt{xy}}(x\frac{dy}{dx}+y)=2\frac{dy}{dx}

\frac{1}{2}\sqrt{\frac{x}{y}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{y}{x}}=2\frac{dy}{dx}

\frac{dy}{dx}=\sqrt{\frac{y}{x}}\cdot\frac{1}{4-\sqrt{\frac{x}{y}}}

Thus, at (-\frac{1}{3},-3),

\frac{dy}{dx}=\sqrt{\frac{-3}{-1/3}}\cdot\frac{1}{4-\sqrt{\frac{-1/3}{-3}}}=\frac{9}{11}

Student 2 gave

\sqrt{x}\cdot \frac{1}{2\sqrt{y}}\cdot \frac{dy}{dx}+\sqrt{y}\cdot \frac{1}{2\sqrt{x}}=2\frac{dy}{dx}

\frac{1}{2}\sqrt{\frac{x}{y}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{y}{x}}=2\frac{dy}{dx}

Thus, at (-\frac{1}{3},-3),

\frac{1}{2}\sqrt{\frac{-1/3}{-3}}\frac{dy}{dx}+\frac{1}{2}\sqrt{\frac{-3}{-1/3}}=2\frac{dy}{dx}

\frac{dy}{dx}=\frac{9}{11}

The answer obtained above is wrong.

By using Desmos, the shape of the equation \sqrt{xy}=7+2y can be easily seen below,

thus \frac{dy}{dx} at (-\frac{1}{3},-3) should be negative, in fact, the correct answer is -\frac{9}{13}.

Students, what’s wrong there?

Think about that:

1. Is it always true that \frac{x}{\sqrt{xy}}=\sqrt{\frac{x}{y}} ?

2. Is it always true that \sqrt{xy}=\sqrt{x}\sqrt{y} ?

Also read
https://johnmayhk.wordpress.com/2018/03/21/just-a-question-about-differentiation/

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1 則迴響 »

  1. Good example. Note that there is a typo in the first line of Student 1’s answer, a factor 1/2 is missing.

    迴響 由 Wilson Cheung Wai-Shun — 2019/05/05 @ 9:55 下午 | 回應


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