Using parametric differentiation for $\frac{\operatorname d \! y}{\operatorname d \!x}$?

Question

Hi so I'm in my calculus class and the teacher gave us a problem to do. I'm not quite sure how to attack this question. He's given us a couple of steps but I don't understand. If someone can further explain to me, I'd really appreciate it.

Question:

Use: $$ x = \frac{2t}{(1+t^2)} \\ y = \frac{t^2}{(1+t^2)} $$

1. Eliminate t from the equations.

2. Find $\frac{\operatorname d \! y}{\operatorname d \!x}$ in terms of x and y.

3. Express your answer back in terms of $t$.

4. Use parametric differentiation for $\frac{\operatorname d \! y}{\operatorname d \!x}$.

I've actually attempted it, I'm not trying to be spoon fed, I just need some further clarification. Examples or sources would be very helpful.

Thanks very much!

Edit:

I've found $\frac{\operatorname d \! y}{\operatorname d \!x}$ in terms of t now:

$\frac{\operatorname d \! y}{\operatorname d \!x}$ = $\frac{\operatorname t \! }{\operatorname -1-t^2 \!}$

But how would I find $\frac{\operatorname d \! y}{\operatorname d \!x}$ in terms of x and y?

Answer 1

This exercise demonstrates the power of parametric differentiation.

The first part wants you to eliminate $t$ from the equations. Your idea is correct. Solve for $t$ in terms of $x$ and plug it back into $y$, so that $y$ is a function of $x$. Then it's quite easy to find $\frac{dy}{dx}$. By the quadratic formula, $$t = \frac{1\pm\sqrt{1-x^2}}{x}.$$ Then $$y = \frac{t^2}{1+t^2} = \frac{\left(\frac{1\pm\sqrt{1-x^2}}{x}\right)^2}{1+\left(\frac{1\pm\sqrt{1-x^2}}{x}\right)^2}.$$ After much work, we find that $$\frac{dy}{dx}=\mp\frac{x}{2\sqrt{1-x^2}}.$$ Now we want $\frac{dy}{dx}$ in terms of $t$, so we plug back in the original $x$. After more work, we have $$\frac{dy}{dx}=\mp\frac{\left(\frac{2t}{1+t^2}\right)}{2\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}} = \mp \frac{t}{t^2-1}.$$ Cool, we're done. But this could have been avoided, had we used the chain rule. Namely, $$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} \implies \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx},$$by dividing both sides by $\frac{dx}{dt}$. Had we simply evaluated $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, we wouldn't have to go through the strenuous algebra overkill above.

Answer 2

From the chain rule,

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

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

Answer 3

Using the chain rule, you know that $\frac{dy}{dt}=\frac{dy}{dx}\left(\frac{dx}{dt}\right)\implies \frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$ This means you need to find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ then you will be able to find $\frac{dy}{dx}.$

So you need to find $$x=\frac{2t}{1+t^2}\implies \frac{dx}{dt}=?$$ and $$y=\frac{t^2}{1+t^2}\implies \frac{dy}{dt}=?$$ $$\implies \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

Answer 4

You need to first find $\frac{\mathrm{d} x}{\mathrm{d} t}$ and $\frac{\mathrm{d} y}{\mathrm{d} t}$.

It follows from the chain rule that $\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}$

$\because 1= \frac{\mathrm{d} y}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} y}$

After computing $\frac{\mathrm{d} x}{\mathrm{d} t}$ and $\frac{\mathrm{d} y}{\mathrm{d} t}$, simply substitute them in to discover $\frac{\mathrm{d} y}{\mathrm{d} x}$ in terms of t.