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I'm having a problem with my solution to a textbook exercise:

Find the Cartesian equation of the curve given by this parametric equation: $$x = \frac{t}{2t-1}, y = \frac{t}{t+1}$$

The textbook's answer is $y = \frac{x}{3x-1}$

My solution is different. I understand how the textbook got to its solution, but I can't find where I made my mistake. Can anyone spot my error below?

$$x = \frac{t}{2t-1} = \frac{t}{2t} - \frac{t}{1}$$ $$\implies x = \frac{1}{2} - t$$ $$\implies x - \frac{1}{2} = -t$$ $$\implies t = -x + \frac{1}{2}$$

Sub this into $y = \frac{t}{t+1} \implies y = \frac{-x + \frac{1}{2}}{-x + \frac{1}{2} + 1}$ $$= \frac{-x + \frac{1}{2}}{-x + \frac{1}{2}} + \frac{-x + \frac{1}{2}}{1}$$ $$= 1 - x + \frac{1}{2}$$ $$= -x + \frac{3}{2}$$

So $y = -x + \frac{3}{2}$


I suspect my error is when I split my fractions up, but if so, why can't I do it like that?

Many thanks!

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Also, is my use of the implication symbol acceptable as written above? Thanks! –  Danny King May 3 '11 at 8:56
4  
$$x = \frac{t}{2t-1} = \frac{t}{2t} - \frac{t}{1}$$, really? –  Raskolnikov May 3 '11 at 8:58
    
Yeah... $x = \frac{t}{2t-1} \neq \frac{t}{2t} - \frac{t}{1}$. There's your error. –  Stijn May 3 '11 at 8:59
2  
Just to clarify, I had mistakenly thought $\frac{a}{b+c} = \frac{a}{b} + \frac{a}{c}$. I was confusing it with the rule $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$ (which I hope is true!) –  Danny King May 3 '11 at 9:10
2  
Danny: you made the mistake of splitting the denominator twice in your derivation: be sure to avoid it by remembering some examples, e.g.: $0.5 = \frac{1}{2} = \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2$. Your second rule in the last comment is perfectly ok. Moreover, your use of the implication symbol looks fine to me. –  t.b. May 3 '11 at 9:12

1 Answer 1

up vote 2 down vote accepted

As you said Danny,

$$\frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$$

and Theo pointed out with a simple example why:

$$0.5 = \frac{1}{2} = \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \; .$$

It is a common mistake however, so tempting that few people have resisted making it.

P.S: Note you made the mistake twice, once in the formula with $x$ and once in the one with $y$.

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