# numeric solving vs graph - Hewlett Packard

This is a discussion on numeric solving vs graph - Hewlett Packard ; Hello all, I had a quite simple problem, or so I thought. I needed to find the first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t). I put the equation on the stack, LS+Calc ...

# Thread: numeric solving vs graph

1. ## numeric solving vs graph

Hello all,

I had a quite simple problem, or so I thought. I needed to find the
first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
formula filling up the display. Functions in ALG don't really make it
look any better, so I leave it as is.
I put 0 on the stack and then =, which gives me dx/dt = 0.

Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
seconds is { }.

Graphing that function, showing f', and getting the roots in Graph
mode reveal the correct answer of t=0.416 for the first, and from that
I can deduce t(n) = 0.416+n*PI/3.

How can I make it work numerically?

Thanks! Hope this was clear enough....

Christoph

2. ## Re: numeric solving vs graph

On Sep 20, 11:11*pm, Christoph Koehler
wrote:
> Hello all,
>
> I had a quite simple problem, or so I thought. I needed to find the
> first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).
>
> I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
> formula filling up the display. Functions in ALG don't really make it
> look any better, so I leave it as is.
> I put 0 on the stack and then =, which gives me dx/dt = 0.
>
> Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
> seconds is { }.
>
> Graphing that function, showing f', and getting the roots in Graph
> mode reveal the correct answer of t=0.416 for the first, and from that
> I can deduce t(n) = 0.416+n*PI/3.
>
> How can I make it work numerically?
>
> Thanks! Hope this was clear enough....
>
> Christoph

I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

The symbolic solver also gave me { }, so I used the numeric solver.
Use STEQ to store level 1 of the stack into variable EQ, then launch
the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
initial guess (I used 1) and I got the root 1.0646... Different
guesses will result in different roots, though this method is somewhat
blind.

If you want the first 3 values where dx/dt = 0, it's probably easier
to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
the graphing environment, move the cursor close to the root, press
FCN, then ROOT. Do this for the first 3 intersections and get t =
1.0646, t = 2.1118, t = 3.1590.

S.C.

3. ## Re: numeric solving vs graph

In article
,
sc_usenet@hotmail.com wrote:

> On Sep 20, 11:11*pm, Christoph Koehler
> wrote:
> > Hello all,
> >
> > I had a quite simple problem, or so I thought. I needed to find the
> > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).
> >
> > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
> > formula filling up the display. Functions in ALG don't really make it
> > look any better, so I leave it as is.
> > I put 0 on the stack and then =, which gives me dx/dt = 0.
> >
> > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
> > seconds is { }.
> >
> > Graphing that function, showing f', and getting the roots in Graph
> > mode reveal the correct answer of t=0.416 for the first, and from that
> > I can deduce t(n) = 0.416+n*PI/3.
> >
> > How can I make it work numerically?
> >
> > Thanks! Hope this was clear enough....
> >
> > Christoph

>
> I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.
>
> The symbolic solver also gave me { }, so I used the numeric solver.
> Use STEQ to store level 1 of the stack into variable EQ, then launch
> the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> initial guess (I used 1) and I got the root 1.0646... Different
> guesses will result in different roots, though this method is somewhat
> blind.
>
> If you want the first 3 values where dx/dt = 0, it's probably easier
> to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> the graphing environment, move the cursor close to the root, press
> FCN, then ROOT. Do this for the first 3 intersections and get t =
> 1.0646, t = 2.1118, t = 3.1590.
>
> S.C.

Curious. For the same dy/dx as you have, I find different zeroes.

Namely: 0.4163, 1.4635 and 2.5101, to 4 decimal places.

4. ## Re: numeric solving vs graph

On Sun, 21 Sep 2008 09:40:12 -0500:

> Different guesses will result in different roots,
> though this method is somewhat blind.

An aid to systematic searching for numeric ROOTS over an interval:

UserRPL program listing:

[r->] [OFF]

5. ## Re: numeric solving vs graph

On Sep 21, 2:13*pm, Virgil wrote:
> In article
> ,
>
>
>
>
>
> *sc_use...@hotmail.com wrote:
> > On Sep 20, 11:11*pm, Christoph Koehler
> > wrote:
> > > Hello all,

>
> > > I had a quite simple problem, or so I thought. I needed to find the
> > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

>
> > > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
> > > formula filling up the display. Functions in ALG don't really make it
> > > look any better, so I leave it as is.
> > > I put 0 on the stack and then =, which gives me dx/dt = 0.

>
> > > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
> > > seconds is { }.

>
> > > Graphing that function, showing f', and getting the roots in Graph
> > > mode reveal the correct answer of t=0.416 for the first, and from that
> > > I can deduce t(n) = 0.416+n*PI/3.

>
> > > How can I make it work numerically?

>
> > > Thanks! Hope this was clear enough....

>
> > > Christoph

>
> > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

>
> > The symbolic solver also gave me { }, so I used the numeric solver.
> > Use STEQ to store level 1 of the stack into variable EQ, then launch
> > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> > initial guess (I used 1) and I got the root 1.0646... Different
> > guesses will result in different roots, though this method is somewhat
> > blind.

>
> > If you want the first 3 values where dx/dt = 0, it's probably easier
> > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> > the graphing environment, move the cursor close to the root, press
> > FCN, then ROOT. Do this for the first 3 intersections and get t =
> > 1.0646, t = 2.1118, t = 3.1590.

>
> > S.C.

>
> Curious. For the same dy/dx as you have, I find different zeroes.
>
> Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.-

That's weird. I just did the same thing again and got your roots. Now
I'm not sure what happened last time.

S.C.

6. ## Re: numeric solving vs graph

sc_usenet@hotmail.com wrote:

> I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.
>
> The symbolic solver also gave me { }, so I used the numeric solver.
> ...
> If you want the first 3 values where dx/dt = 0, it's probably easier
> to use the grapher.
> ...
> Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> the graphing environment, move the cursor close to the root, press
> FCN, then ROOT.

Well, I don't want to disturb this discussion about various more or less
elegant ways to make the calculator solve the problem, but in this case
there is a much easier and more straightforward way: simply use your own
brain, the calculator almost provided the solution anyway:

> I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

So the rest is trivial: a local minimum or maximum of x(t) requires
dx/dt = 0 => 12 cos(3t) - 4 sin(3t) = 0 => t = 1/3 arctan 3 = 0.4163...

Since tan x = tan(x + Pi), simply add multiples of Pi/3 for further
solutions, i.e. t = 0.4163, 1.4635, 2.5107, 3.5579, 4.6051...

It's that easy.
Okay, I didn't check the second derivate. <8)

Maybe this is a good example for combining the benefits of our little
electronic helpers with good old plain man-made math. The former
provides the derivate, the latter gives the final result.
In this case even the correct one. 8-)

Dieter

7. ## Re: numeric solving vs graph

On Sep 21, 8:07*pm, Dieter wrote:
> Maybe this is a good example for combining the benefits of our little
> electronic helpers with good old plain man-made math. The former
> provides the derivate, the latter gives the final result.
> In this case even the correct one. 8-)
>

Exactly -- the calculator is (or should be) and extension to the
brain, not a replacement.

I didn't think about that though, since I was more focused on
calculator usage with this question.

S.C.

8. ## Re: numeric solving vs graph

In article
,
sc_usenet@hotmail.com wrote:

> On Sep 21, 2:13*pm, Virgil wrote:
> > In article
> > ,
> >
> >
> >
> >
> >
> > *sc_use...@hotmail.com wrote:
> > > On Sep 20, 11:11*pm, Christoph Koehler
> > > wrote:
> > > > Hello all,

> >
> > > > I had a quite simple problem, or so I thought. I needed to find the
> > > > first 3 values of t for which dx/dt =0, with x(t) = 4 sin3t e^(-t).

> >
> > > > I put the equation on the stack, LS+Calc -> DERVX. It's some crazy
> > > > formula filling up the display. Functions in ALG don't really make it
> > > > look any better, so I leave it as is.
> > > > I put 0 on the stack and then =, which gives me dx/dt = 0.

> >
> > > > Then I go to LS+S.SLV -> SOLVE (the first one). The result after a few
> > > > seconds is { }.

> >
> > > > Graphing that function, showing f', and getting the roots in Graph
> > > > mode reveal the correct answer of t=0.416 for the first, and from that
> > > > I can deduce t(n) = 0.416+n*PI/3.

> >
> > > > How can I make it work numerically?

> >
> > > > Thanks! Hope this was clear enough....

> >
> > > > Christoph

> >
> > > I got that dx/dt = (12 cos(3t) - 4 sin(3t))/e^t.

> >
> > > The symbolic solver also gave me { }, so I used the numeric solver.
> > > Use STEQ to store level 1 of the stack into variable EQ, then launch
> > > the numeric solver ([RS]7 (NUM.SLV) 1. Solve equation). Supply an
> > > initial guess (I used 1) and I got the root 1.0646... Different
> > > guesses will result in different roots, though this method is somewhat
> > > blind.

> >
> > > If you want the first 3 values where dx/dt = 0, it's probably easier
> > > to use the grapher. Use STEQ, then hold [LS] F1, ERASE DRAW. When in
> > > the graphing environment, move the cursor close to the root, press
> > > FCN, then ROOT. Do this for the first 3 intersections and get t =
> > > 1.0646, t = 2.1118, t = 3.1590.

> >
> > > S.C.

> >
> > Curious. For the same dy/dx as you have, I find different zeroes.
> >
> > Namely: 0.4163, 1.4635 and 2.5107, to 4 decimal places.-

>
> That's weird. I just did the same thing again and got your roots. Now
> I'm not sure what happened last time.
>
> S.C.

We'll probably never know.

9. ## Re: numeric solving vs graph

On Sep 21, 7:07*pm, Dieter wrote:
> Maybe this is a good example for combining the benefits of our little
> electronic helpers with good old plain man-made math. The former
> provides the derivate, the latter gives the final result.
> In this case even the correct one. 8-)

I totally agree.
I can do this by hand. But in a test setting I'd rather concentrate on
the subject matter and not worry about taking derivatives. This is not
for a Calculus class but EE.

Thanks for all the tips

Christoph