Conversion from Parametric to Cartesian

• W3bbo
Thus, \sqrt{\frac{y}{2}} = \pm\sqrt{1 + \left(\frac{1}{2}x\right)^2}This is the same equation as above, since it's only the \pm sign that's different.Edit: Whoops, I see you've already gotten it. Sorry for the redundant post.
W3bbo

Homework Statement

Reduce these parametric functions to a single cartesian equation:

$$\displaylines{ x = at^2 \cr y = 2at \cr}  \displaylines{ x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr}  \displaylines{ x = t^2 + 1 \cr y = t^2 + t \cr}$$

N/A

The Attempt at a Solution

For Q1, I think I got the hang of it... in that the plots of my cartesian and the paremetric are identical:

$$\displaylines{ x = at^2 \cr y = 2at \cr \cr t^2 = {x \over a} \cr t = \sqrt {{x \over a}} \cr \cr y = 2a\sqrt {{x \over a}} \cr}$$

For Q2 I thought I had it solved with a trigonometric identity, but the plots look different:

$$\displaylines{ x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr \cr {\mathop{\rm Sec}\nolimits} ^2 \left( \alpha \right) = 1 + {\mathop{\rm Tan}\nolimits} ^2 \left( \alpha \right) \cr \left( {{\textstyle{1 \over 3}}x} \right)^2 = 1 + \left( {{\textstyle{1 \over 5}}y} \right)^2 \cr {\textstyle{1 \over 9}}x^2 = 1 + {\textstyle{1 \over {25}}}y^2 \cr {\textstyle{1 \over {25}}}y^2 = {\textstyle{1 \over 9}}x^2 - 1 \cr y^2 = {\textstyle{{25} \over 9}}x^2 - 1 \cr y = \pm \sqrt {{\textstyle{{25} \over 9}}x^2 - 1} \cr}$$

Finally, Q3 seems deceptivly simple, but again, the plot doesn't match the original:

$$\displaylines{ x = t^2 + 1 \cr y = t^2 + t \cr \cr t^2 = x - 1 \cr y = x - 1 + \sqrt {x - 1} \cr}$$

I'm not looking for answers, just to find out where I've gone wrong.

Thanks

Q2. You've made an error when multiplying through by 25.. you should have a -25 on the RHS of the penultimate line.

Q3. I'm not sure what you've done, but I'd try solving one of the parametric equations for t, but the quadratic formula, then subbing into the other equation.

Thanks. I got those questions correct.

Further on I'm faced with converting this into cartesian:
$$\displaylines{ x = 2{\mathop{\rm Cot}\nolimits} \left( t \right) \cr y = 2{\mathop{\rm Sin}\nolimits} ^2 \left( t \right) \cr}$$​

I tried having a go using this:

$$\displaylines{ x = 2{\mathop{\rm Cot}\nolimits} \left( t \right) \cr t = {\mathop{\rm Tan}\nolimits} ^{ - 1} \left( {{\textstyle{2 \over x}}} \right) \cr y = 2{\mathop{\rm Sin}\nolimits} ^2 \left( {{\mathop{\rm Tan}\nolimits} ^{ - 1} \left( {{\textstyle{2 \over x}}} \right)} \right) \cr}$$​

...but the plot is totally different.

I was able to solve it eventually using

$$\displaylines{ 1 + {\mathop{\rm Cot}\nolimits} ^2 \left( t \right) = {\mathop{\rm Cosec}\nolimits} ^2 \left( t \right) \cr 1 + \left( {{\textstyle{1 \over 2}}x} \right)^2 = \left( {{\textstyle{y \over 2}}} \right)^{ - 1} \cr} p$$​

But can you explain why my initial attempt failed?

Thanks

Well, when you divide by 2, you should have x/2, not 2/x

Tedjn said:
Well, when you divide by 2, you should have x/2, not 2/x

Note it's Cot, not Tan. So I inverted the argument for Tan^-1. Otherwise it would have been ArcCot(x/2)

Sorry, I see that now. It's probably some quirk of the graphing software. The first set should work out to the second set if you get rid of the inverse trig functions.

$$y = 2\sin^2\left[\tan^{-1}\left(\frac{2}{x}\right)\right]$$

From a picture of the right triangle, we know that

$$\sin t = \pm\frac{2}{\sqrt{4+x^2}} \implies \sin^2 t = \frac{4}{4 + x^2}$$

Then,

$$\frac{y}{2} = \frac{4}{4+x^2}\longrightarrow \left(\frac{y}{2}\right)^{-1} = \frac{4+x^2}{4} = 1 + \frac{x^2}{4} = 1 + \left(\frac{1}{2}x\right)^2$$

What is the purpose of converting from parametric to Cartesian form?

The purpose of converting from parametric to Cartesian form is to express a set of parametric equations in terms of x and y, making it easier to graph and analyze the equations. Cartesian form also allows for easier manipulation and calculation of the equations.

How do you convert from parametric to Cartesian form?

To convert from parametric to Cartesian form, you must first eliminate the parameter. This can be done by solving one of the equations for the parameter and substituting it into the other equation. Then, simplify the equation to get it in the form of y = f(x).

Can all parametric equations be converted to Cartesian form?

No, not all parametric equations can be converted to Cartesian form. In some cases, the equations may be too complex or involve trigonometric functions that cannot be easily eliminated. In these cases, it may be more beneficial to work with the parametric form of the equations.

What are the benefits of using parametric form over Cartesian form?

One benefit of using parametric form is that it allows for more flexibility in representing curves and shapes. Parametric equations can also be useful in solving problems involving motion or changing variables. Additionally, some curves may be easier to graph and analyze in parametric form rather than Cartesian form.

How can I check if my conversion from parametric to Cartesian is correct?

To check if your conversion is correct, you can graph both the parametric and Cartesian forms of the equations and see if they produce the same curve. You can also plug in different values for x and y to see if they satisfy both equations. Additionally, you can use a graphing calculator or software to verify the accuracy of your conversion.

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