Normal to a space curve.

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March 17th 2009, 04:17 AM
scorpion007

Normal to a space curve.

The question is:

Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
$\mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}$
at the point P(32, -32, 16).
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Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via $\mathbf{n} \cdot \mathbf{v} = 0$ where $\mathbf{v} = \mathbf{Q}-\mathbf{P}$ (the vector from the plane's origin (P) to any point Q).

Of course, I can find the derivative of r(t), which is:

$\dot{\mathbf{r}}(t) = 4t^{-3/4}\mathbf{i}-3t^{-1/4}\mathbf{j}+\mathbf{k}$

That will give me the tangent vector at any time t.

I know that t = 16, from the 'k' component, so I can evaluate $\dot{\mathbf{r}}(16) = (\frac{1}{2}, -\frac{3}{2}, 1)$

But I need the normal vector (perpendicular to the curve).

Any tips?
March 17th 2009, 04:36 AM
Mush

Quote:

Originally Posted by scorpion007

The question is:

Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
$\mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}$
at the point P(32, -32, 16).
-----------

Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via $\mathbf{n} \cdot \mathbf{v} = 0$ where $\mathbf{v} = \mathbf{Q}-\mathbf{P}$ (the vector from the plane's origin (P) to any point Q).

Of course, I can find the derivative of r(t), which is:

$\dot{\mathbf{r}}(t) = 4t^{-3/4}\mathbf{i}-3t^{-1/4}\mathbf{j}+\mathbf{k}$

That will give me the tangent vector at any time t.

I know that t = 16, from the 'k' component, so I can evaluate $\dot{\mathbf{r}}(16) = (\frac{1}{2}, -\frac{3}{2}, 1)$

But I need the normal vector (perpendicular to the curve).

Any tips?

A unit normal of a surface S, at point P is:

Try this:

$\frac{\vec{u}\text{ }'(t)}{|\vec{u}\text{ }'(t)|}$

Where $\vec{u(t)}$ is the unit tangent vector.

If you have found the tangent vector, then to get the unit tangent vector you must divide the tangent vector by it's magnitude.
March 17th 2009, 04:47 AM
scorpion007

So if I understand correctly, u(t) = r'(t)/|r'(t)| ? (Which is the unit tangent vector at 't').

Then u'(t) = r''(t)? (The derivative of r'(t)).

So you want me to find, $\frac{\ddot{\mathbf{r}}(t)}{|\ddot{\mathbf{r}}(t)| }$ ?

EDIT:

Yes, I know how to find a unit vector from a vector, but I'm interested in the perpendicular vector to the tangent, not the tangent.
March 17th 2009, 04:50 AM
Mush

Quote:

Originally Posted by scorpion007

So if I understand correctly, u(s) = r'(t)/|r'(t)| ? (Which is the unit tangent vector at 't').

Then u'(t) = r''(t)? (The derivative of r'(t)).

So you want me to find, $\frac{\ddot{\mathbf{r}}(t)}{|\ddot{\mathbf{r}}(t)| }$ ?

EDIT:

Yes, I know how to find a unit vector from a vector, but I'm interested in the perpendicular vector to the tangent, not the tangent.

I know what you're trying to find. But in order to find the unit normal vector, you must find the unit tangent vector, as is shown below:

$\vec{n} (t) = \frac{\vec{u}\text{ } ' (t)}{|\vec{u}\text{ } '(t)|}$

Where

$\vec{u} (t) = \frac{\vec{r}\text{ }'(t)}{|\vec{r}\text{ }'(t)|}$
March 17th 2009, 05:59 AM
scorpion007

Ok, so the unit tangent vector is simply $\frac{\sqrt{7}}{\sqrt{2}}(1/2,-3/2,1)$ , right?
March 17th 2009, 04:54 PM
scorpion007

Hmmm.... all the expressions I get are really complex. The book has the answer (for the plane equation) that is nice and simple looking (no square roots, or fancy powers, or fractions, or anything like that!):

$x-3y+2z=160$

Could someone explain how they got that?
March 17th 2009, 05:42 PM
Reckoner

You do not need a vector that is normal to the curve (such a vector would be of no use in this problem). You only need one that is normal to the desired plane.

Quote:

Originally Posted by scorpion007

Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
$\mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}$
at the point P(32, -32, 16).

$\mathbf r'$ is tangent to the curve, so we find it to be

$\mathbf r'(t)=4t^{-3/4}\mathbf i-3t^{-1/4}\mathbf j+\mathbf k.$

The point $P(32,\,-32,\,16)$ corresponds to $t=16$ and

$\mathbf r'(16)=\frac12\mathbf i-\frac32\mathbf j+\mathbf k.$

This vector is tangent to the curve, so it must be normal to the plane perpendicular to the curve. Using the point $(32,\,-32,\,16)$ and normal vector $2\mathbf r'(16),$ the equation of the plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

$\Rightarrow (x-32)-3(y+32)+2(z-16)=0$

$\Rightarrow x-3y+2z-160=0.$
March 17th 2009, 08:30 PM
scorpion007

Oh my! Thank you so much! It was way easier than I thought. I misunderstood the question initially, as you can see.

I didn't realise that the derivative of r(t) was in fact going to give me normal vectors for planes at each 't' that are perpendicular to r(t). I see that now, thanks to you.

Once I understood that, the rest is easy!