Convert f(x) to parametric holding something constant

[ChickenChakuro]

Hi all, if I have a problem like:

The path of a particle is described by $$y=4x^2$$, and it has a constant velocity of 5 m/s.

How do I make a parametric equation out of this?

I tried doing:
$$r = xi + f(x)j$$
$$r = ti + 4t^2j$$,

but then $$v = \frac{dr}{dt} = i + 8tj$$, so $$|v|=\sqrt{1+64t^2}$$, and at, for example, time $$t = 0$$, $$|v| = \sqrt{1+0} = 1$$, not the constant 5 m/s! Anyone know how I can make a parametric equation that will work?

 

[Zurtex]

So you have y = 4x² and |v(t)| = 5 (and v(t) > 0)

Letting x = (x(t),y(t))

You know that dv/dt = x

So you know that:

| integral(x,dt) | = 5

Think about it for a bit, sure it will come to you

 

[ChickenChakuro]

I don't get it!

I tried:

$$|(\int x\,dt)| = 5$$

$$\sqrt{(xt)^2+C^2}= 5$$

$$x^2t^2+C^2=25$$

It doesn't seem to work! When x = 0, C=5, so
$$|v| = x^2t^2 +5$$?

 

[HallsofIvy]

Then length is NOT $| \int x dx|$! Also, x is itself a function of t so your integration is wrong.

If x= f(t) and y= g(t) then since y= 4x², g(t)= 4(f(t)² so
$$\frac{dg}{dt}= 8f(t)\frac{df}{dt}$$
The length f(t)i+ g(t)j is
$$\sqrt{f^2(t)+ 64f^2(t)\left(\frac{df}{dt}\right)^2}= 5$$
Solve that differential equation for f(x) and then find g(x).

 

[ChickenChakuro]

Oh, thanks. Wait, how do I solve for f(x)? All I see is f(t)!

 

[ChickenChakuro]

Ah, nevermind. Thanks for the help!

 

[nvn]

There appear to be typographic mistakes in post 4. For example,

$$\sqrt{f^2(t)+64f^2(t)\left(\frac{df}{dt}\right)^2}\;=\:5\quad\ \ \mbox{should instead be}$$

$$\sqrt{\left(\frac{df}{dt}\right)^2+64f^2(t)\left(\frac{df}{dt}\right)^2}=\:5.$$​

I would instead solve the given problem in post 1 as follows.

Given y = 4*x², ds/dt = 5 m/s. At x = 0, t = 0.

$$\mathbf{\vec s}=x\,\mathbf{i}+y\,\mathbf{j}$$

$$d\mathbf{\vec s}=dx\,\mathbf{i}+dy\,\mathbf{j}$$

$$ds=[(dx)^2+(dy)^2]^{0.5}$$​

$$\mbox{But since }y=4x^2,\ \ \therefore \ dy=8x\,dx.\ \mbox{Therefore,}$$

$$\begin{equation*}\begin{split}ds&=[(dx)^2+64x^2(dx)^2]^{0.5}\\<br /> &=(1+64x^2)^{0.5}\,dx\end{equation*}\end{split}$$

$$\frac{ds}{dt}=(1+64x^2)^{0.5}\,\frac{dx}{dt}$$​

$$\mbox{But }\frac{ds}{dt}=5\ \mbox{m/s}.$$

$$5=(1+64x^2)^{0.5}\,\frac{dx}{dt}$$

$$\begin{equation*}\begin{split}dt&=\frac{1}{5}(1+64x^2)^{0.5}\,dx\\<br /> &=\frac{8}{5}(\frac{1}{64}+x^2)^{0.5}\,dx\\<br /> &=\frac{8}{5}[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx\end{equation*}\end{split}$$

$$\int dt=\frac{8}{5}\int[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx$$

$$t=\frac{8}{5}\int[\left(\frac{1}{8}\right)^2+x^2]^{0.5}\,dx$$​

Therefore, from any integral table,

$$t\:=\:\frac{4}{5}x\,(\frac{1}{64}+x^2)^{0.5}\ +\ \frac{1}{128}\ln|x+(\frac{1}{64}+x^2)^{0.5}|\ +\ c$$​

$$\mbox{At}\ x=0,\ t=0;\ \ \therefore \ c=-\frac{1}{128}\ln(\frac{1}{8}).$$

$$\mbox{Now, let }\:x=\frac{1}{8}\sinh z.\ \ \mbox{Therefore,}$$

$$\begin{equation*}\begin{split}t\:&=\:\frac{1}{10}(\sinh z)(\frac{1}{64}+\frac{1}{64}\sinh^2z)^{0.5}\ \\<br /> &+\ \frac{1}{128}\ln\left[\frac{1}{8}\sinh z+(\frac{1}{64}+\frac{1}{64}\sinh^2z)^{0.5}\right]\ -\ \frac{1}{128}\ln(\frac{1}{8})\\<br /> &=\:\frac{1}{80}(\sinh z)(1+\sinh^2z)^{0.5}\ \\<br /> &+\ \frac{1}{128}\ln\left(\frac{1}{8}[\sinh z+(1+\sinh^2z)^{0.5}]\right)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\<br /> &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln\left[\frac{1}{8}(\sinh z+\cosh z)\right]\ -\ \frac{1}{128}\ln(\frac{1}{8})\\<br /> &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln(\frac{1}{8}\,e^z)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\<br /> &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\ln(\frac{1}{8})\ +\ \frac{1}{128}\ln(e^z)\ -\ \frac{1}{128}\ln(\frac{1}{8})\\<br /> &=\:\frac{1}{80}(\sinh z)(\cosh z)\ +\ \frac{1}{128}\,z\\<br /> &=\:\frac{1}{160}\sinh(2z)\;+\;\frac{1}{128}\,z\end{split}\end{equation*}$$​

Unfortunately, this equation cannot be solved analytically for z (to my knowledge). But you could solve it for z numerically, for any given value of t. And then you can compute the corresponding value of x and y, because x = 0.125*sinh(z), and y = 4*x².

 

[HallsofIvy]

Thanks, but this thread was 4 years old!