1. ## Parametric Differentiation

Q:
The parametric equations of a curve are x = a cos^3 (n) , y = a sin^3 (n), where a is a positive constant and 0 < n < (pi/2). The tangent to the curve at the point with parameter n cuts the x - and y - axes at S and T respectively. Find the equation of this tangent, and show that the distance ST is independent of n.

Thank you for helping!

2. Hello,

Originally Posted by Tangera
Q:
The parametric equations of a curve are x = a cos^3 (n) , y = a sin^3 (n), where a is a positive constant and 0 < n < (pi/2). The tangent to the curve at the point with parameter n cuts the x - and y - axes at S and T respectively. Find the equation of this tangent, and show that the distance ST is independent of n.

Thank you for helping!
According to your title, what you need is this :

$\frac{dy}{dx}=\frac{dy}{dn} \cdot \frac{dn}{dx}=\frac{dy}{dn} \cdot \frac{1}{\frac{dx}{dn}}$

isn't it ?

3. Hello Moo!
I found dy/dx, but I am not sure how to continue to find the equation of the tangent...

4. Originally Posted by Tangera
Hello ...!
I found dy/dx, but I am not sure how to continue to find the equation of the tangent...
I didn't do the next steps - but I would do the following:

1. Choose an arbitrary point $Q(a(\cos(q))^3~,~a(\sin(q))^3)$

2. Calculate the equation of the tangent to the curve in Q. Use point-slope-formula of a straight line.

3. Calculate the coordinates of the x- and y-intercepts.

4. Calculate the distance between these intercepts. It is sufficient to calculate $d^2$ instead of d. You should come out with a term only depending on a, that means without any q left.

5. Originally Posted by Tangera
Q:
The parametric equations of a curve are x = a cos^3 (n) , y = a sin^3 (n), where a is a positive constant and 0 < n < (pi/2). The tangent to the curve at the point with parameter n cuts the x - and y - axes at S and T respectively. Find the equation of this tangent, and show that the distance ST is independent of n.

Thank you for helping!
We know that the equation of a tangent line is

$y-f(x_0)=f'(x_0)(x-x_0)$

Now lets think, we want the y coordinate ( $f(x_0)$) at whatever point we are looking to find the tangent line of.

But we know when parametrized we will be given the coordinates in terms of the parametr and not the xy plane.

So we need to find the y coordinate at $t=?$ where t is the parameter.

Well this is pretty easy since we are given that

$y=y(t)$

More specifically you are given that

$y=a\sin^3(t)$

So in other words at lets say $t=\square$

we have that the y coordinate is

$y=a\sin^3\left(\square\right)$

So now we can see that given a t value for the parameter we can start to rewrite out tangent line equation

$y-y(t)=\cdots$

Now next as Moo showed we need to compute

$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$

So Let $f'(t)=\frac{y'(t)}{x'(t)}$

So we can see that $f'(x)=\frac{y'(t)}{x'(t)}$ where t is the corresponding parameter.

So now we can rewrite our tangent line equation a little more

$y-y(t)=\frac{y'(t)}{x'(t)}\cdots$

Finally seeing that we have that

$x=x(t)$

We can see that at the point $t$ we can just replace x with x(t)

Giving us our grand finale of

$y-y(t)=\frac{y'(t)}{x'(t)}\left(x-x(t)\right)$
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Also note that you can deparametrize this

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=2a^{\frac{2}{3}}$