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junior college 1 | H2 Maths
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Qianqi Zhao
Qianqi Zhao

junior college 1 chevron_right H2 Maths chevron_right Singapore

This is on curve sketching on parametric and Cartesian graphs. How do I change from parametric equation to Cartesian equation when the there is a t^2? How do I eliminate t?

Date Posted: 4 years ago
Views: 540
Eric Nicholas K
Eric Nicholas K
4 years ago
For part ii, you can also find an expression for “dx/dy” from the expression

y^2 = x^2 (x - 3)

and find the “minimum point” of the curve of x in terms of y.

As mentioned earlier, the graph containing a +- (or equivalently, the graph of y^2 = f(x)) is always symmetrical about the x-axis. This is an important property to know.

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Eric Nicholas K
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There are other “more algebraic” methods to do part i and ii, but these approaches are sufficiently algebraic.
Qianqi Zhao
Qianqi Zhao
4 years ago
Thank you so much! I was able to understand part i. However, i dont really understand why for part ii, since x-3 is in a square root, x-3 is more than and equal to 0? What about the x outside of the square root?
Eric Nicholas K
Eric Nicholas K
4 years ago
Now, y = x sqrt (x - 3).

The x outside the square root can work for any value of x. However, the one side square root x is only valid for x >= 3, since the square root of any negative number is not a real number. As a result, if x < 3, this expression will not have a real value, even though this fits into the first x.

Points can only exist on the graph if the coordinates have real values.

It’s similar to saying “If one drop of poison is placed in my glass of water, I say that the all the water becomes poisoned and not just a small part of it gets poisoned”. Similarly, the values that x can take are limited to the most restricted expression.

This is why the curve is not defined for x < 3.