Calculating Surface Area Using Double Integrals

[Cyrus]

Hi,

I have a question on the method of calculation of the surface area of a surface. I am using "Calculus Concepts and contexts by stewart", chapter 12.6.

In it, he goes on to explain how to calculate the suface area of a surface as a double integral by using approximations. He breaks up the projection of the surface onto the x-y plane into a grid. And each grid maps out a grid onto the surface as well. Now he wants to find an approximate area of each of those squares, and sum them. A rieman sum basically. Now on the projection he lables each side of the square delta x, and delta y.

He then goes on to say that he needs to approximate each of those patches on the surface, and that can be done if you use two vectors that start at one corner of the patch and both end at each end of the patch. You can then do the cross product of these two vectors to find the area of one patch. Similarly, you can do this for all the patches and sum it.

Now, in order to approximate these vectors, he does the partial with respect to x, to find the tangent to approximate one side of the patch, and similarly with respect to y for the other side. These gives him two tangent vectors that lie on the tangen plane. And he crosses them to approximate the area of the patch.

Here come the confusion. When he calculates the area, he multiplies the tangent with respect to x, by delta x, and similarly for y, by delta y. Why does he multiply the tangents by these values?

Also, he WANTS the two vectors to approximate the sides of the patch, so that the area closely computes the area of the patch; however, the tangent vectors WRT x, and WRT y, can have different magnitudes depending on the origional function that defines the surface. So what makes them work out so that the magnitude will ALWAYS give a good approximation for the area of the patch?

If the patch is very small, then the tangent vectors WRT x and y might be way too big. So when you cross them, you get a huge area relative to what the patch area SHOULD be. So what fixes this dilema?

Thanks for your help all,

Cyrus

 

[Galileo]

Consider a small patch of surface traced out by the vector [itex]\vec r(u,v)[/itex] when u goes from [itex]u_i[/itex] to [itex]u_i+\Delta u[/itex] and v from [itex]v_i[/itex] to [itex]v_i+\Delta v[/itex].
Then the patch can ve approximated by the parallelogram spanned by the vectors [itex]\vec r(u_i+\Delta u_i,v_i)-\vec r(u_i,v_i)[/itex] and [itex]\vec r(u_i,v_i + \Delta v)-\vec r(u_i,v_i)[/itex].
These in turn are approximated by [itex]\vec r_u(u_i,v_i)\Delta u[/itex] and [itex]\vec r_v(u_i,v_i)\Delta v[/itex].
Because clearly:
[tex]\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)[/tex].

That's why the surface area patch is approximated by:
[tex]|(\Delta u \vec r_u(u_i,v_i) \, \times \, \Delta v \vec r_v(u_i,v_i)|=|\vec r_u(u_i,v_i) \times \vec r_v(u_i,v_i)|\Delta u \Delta v[/tex].

Note that it is NOT the projection of the surface onto the xy-plane that is cut into a grids, but the domain D of the function, which is some region in the uv-plane.

For the question about the tangent vectors. It is true that sometimes they may be very big, but that is because of the particle parametrization of the surface that is chosen. Recall that [itex]\vec r_u[/itex] can be considered as the speed at which a curve in the surface is traced out. If the speed is big a greater side of the parallelogram we considered is traced out when u goes a little distance [itex]\Delta u[/itex], so the surface area patch will be greater as well.
You may be more off from the true area of the surface patch when [itex]r_u[/itex] is big, but in the limit as [itex]\Delta u[/itex] goes to zero it approximation becomes exact. Fortunately it can be proven that the surface area is independant from the chosen parametrization.

(The same thing occurs in a standard one dimensional integral. If f(x) varies wildly (almost vertically, thus f'(x) very big), the sum of rectangles approximation gets bad, but in the limit [itex]\Delta x \to 0[/itex] it will still be exact).

 

[Cyrus]

But this: [itex]\vec r(u,v)[/itex] is for a scalar function. How about when it is a vector function: [itex]\vec r(u,v)=x(u,v)i + y(u,v)j + z(u,v)k [/itex]

What meaning does the partial have in this case, becuase it is now a vector function.

 

[Galileo]

I did mean for [itex]\vec r(u,v)[/itex] to be a vector-valued function (hence the arrow on the [itex]\vec r)[/itex].

Suppose you hold v constant and vary u. Then [itex]\vec r(u,v)[/itex] traces out a curve in [itex]\mathbb{R}^3[/itex].
[itex]\vec r_u(u,v)[/itex] simply gives the tangent vector along this curve, just like if you where considering a vectorfunction of one variable [itex]\vec r(t)[/itex]. (It's the rate at which the curve is traced out).
If you vary v and hold u constant, you trace out a different curve ofcourse and [itex]\vec r_v[/itex] is interpreted similarly.

 

[Cyrus]

A question on how you used the following:

[tex]\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)[/tex]

You see, I have seen this notation used on a scalar function but not on a vector function.

 

[dextercioby]

It doesn't really matter whether u meet scalar derivatives,or vector derivatives.Since scalars are vector components in a basis,thing are quite similar,since any vector function's derivative can,intur,be decomposed into scalar components.

Daniel.

PS.Think about physical situations.The simplest:velocity vector is the time derivative of the position vector.

 

[Cyrus]

Ok dex, here's a reworded question. I was not clear becuase I did not understand my question very well myself. Heres what I want to ask.

[tex]\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)[/tex]

We have all this junk up above. Now let's say that we do the partial dervative of a parametric curve

[tex] r(u,v)=x(u,v)i + y(u,v)j+z(u,v)k [/tex]

Now let's say its with respect to u, so that v is held fixed.

Then the partial derivative, (depending on the surface), may have a variable 'u' to some power in it for either the i,j, k or even all directions. I thought that the goal was to find a tangent vector that only had a component in two orthogonal directions. For instance, we want to find a tangent so that it goes delta x in the i direction and delta y in the j direction, but does not change in the k direction. Then we can approximate the area of the patch. But if you look above, it might have the variable 'u' in all three directions, so it might have a change in all three directions. Is this not a problem?

heres my point, one tangent vector should be parallel to the x axis, and the other to the y axis. But this isint true if the partial derivative contains a variable in each direction.

 

[Cyrus]

Take this for example,

[tex]\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)[/tex]

r is a vector function. So would I have to do the following?

[tex]\frac{ x(u_i+\Delta u,v_i)+ x(u_i,v_i)}{\Delta u}i+ \frac{ y(u_i+\Delta u,v_i)-y(u_i,v_i)}{\Delta u}j + \frac{z(u_i+\Delta u,v_i)-z(u_i,v_i)}{\Delta u}k \approx \vec r_u(u_i,v_i)[/tex]

Also, a level curve means that it is horizontal or verticle. But to clearify, this means that the curve is horiztonal or verticle on the domain, right? If you hold u fixed, and vary v, then it may or may not be horizontal or verticel on the SURFACE. That would depenend on how the surface is defined. This sort of anwsers my own question previously, but i don't know if I am wright or not.

 

[Galileo]

Then the partial derivative, (depending on the surface), may have a variable 'u' to some power in it for either the i,j, k or even all directions. I thought that the goal was to find a tangent vector that only had a component in two orthogonal directions. For instance, we want to find a tangent so that it goes delta x in the i direction and delta y in the j direction, but does not change in the k direction. Then we can approximate the area of the patch. But if you look above, it might have the variable 'u' in all three directions, so it might have a change in all three directions. Is this not a problem?

heres my point, one tangent vector should be parallel to the x axis, and the other to the y axis. But this isint true if the partial derivative contains a variable in each direction.

No. It seems like you think the surface is approximated by horizontal patches. This is not the case. It is approximated by the parallelogram spanned by two tangent vectors. One in the direction of [itex]r_u[/itex] and the other in the direction of [itex]r_v[/itex]. The surface patches could even by vertical. Like in the following case of a hollow cylinder of height h and radius R:

[tex]x(u,v)=R\cos(u),\, y(u,v)=R\sin(u), \, z(u,v)=v[/tex]
with [itex]0\leq u < 2\pi[/itex] and [itex]0 \leq v \leq h[/itex].

Then since:
[tex]\vec r_u(u,v)=\frac{\partial x}{\partial u}\vec i+\frac{\partial y}{\partial u}\vec j+\frac{\partial z}{\partial u}\vec k=-R\sin(u)\vec i+R\cos(u)\vec j[/tex]

and

[tex]\vec r_v(u,v)=\frac{\partial x}{\partial v}\vec i+\frac{\partial y}{\partial v}\vec j+\frac{\partial z}{\partial v}\vec k=\vec k[/tex]
we see that [itex]\vec r_v[/itex] points straight up and a surface patch has an area of approximately:
[tex]|\vec r_u \times \vec r_v|\Delta u \Delta v = R\Delta u \Delta v[/tex]

heres my point, one tangent vector should be parallel to the x axis, and the other to the y axis. But this isint true if the partial derivative contains a variable in each direction.

The tangent vectors can have ANY direction whatsoever. Since the surface can be anything at all. Like a sphere. The tangent vectors point in different directions all over the sphere.

If you have the same version of Stewart as I have. Check out paragraph 10.5 about parametric surfaces (Figure 5).

 

[Cyrus]

Ok, just to make sure though, the curve on the domain D; however, will be parallel to the x-axis or y axis, if you take the partial derivative.

 

[Galileo]

Also, a level curve means that it is horizontal or verticle. But to clearify, this means that the curve is horiztonal or verticle on the domain, right? If you hold u fixed, and vary v, then it may or may not be horizontal or verticel on the SURFACE. That would depenend on how the surface is defined. This sort of anwsers my own question previously, but i don't know if I am wright or not.

Sorry, I don't see the connection between the previous and level curves.

Level curves are defined for functions f of two variables.
They are the points in the domain of f for which f(x,y)=k. There's no connection with surface area.

 

[Galileo]

Ok, just to make sure though, the curve on the domain D; however, will be parallel to the x-axis or y axis, if you take the partial derivative.

There is no x- or y-axis for the domain D. Since D exist in the uv-plane.
There are two different sets at work here. [itex]\vec r[/itex] is a function from the uv-plane to [itex]\mathbb{R}^3[/itex], with x,y and z coordinates.

[tex]\vec r : D \to \mathbb{R}^3[/tex]

 

[Cyrus]

Ok, here's my point though, let's say you take the partial derivative of a function with respect to x, then is the tangent vector you get parallel to the x-axis?

P.S. whoops I dident mean level curves, I ment the curve you get when you hold on variable fixed and let the other variable vary, I can't think of what its called at the moment, it slipped my mind.

 

[Galileo]

Plenty of room for confusion here. When you say the partial derivative of a function wrt x, then I assume you are talking about a function of two (or more) variables.
Set z=f(x,y). Then the partial derivative wrt x is not a vector, but a function of x and y.

If you mean [itex]\vec r(u,v)[/itex] then you cannot take the partial derivative with respect to x, because it's a function of u and v.

In light of what you said before:

Also, a level curve means that it is horizontal or verticle. But to clearify, this means that the curve is horiztonal or verticle on the domain, right? If you hold u fixed, and vary v, then it may or may not be horizontal or verticel on the SURFACE. That would depenend on how the surface is defined. This sort of anwsers my own question previously, but i don't know if I am wright or not.

And by level curve you meant the curve traced out when u varies but v is held constant. Then yeah, in the uv-plane, v is constant and u varies, so you're taking a horizontal line in D. And the curve that is traced out are those values of [itex]\vec r(u,v)[/itex] for which u and v lie on the line in D. The curve that is traced out can be anything, and it doesn't have to be horizontal.

Take the example of the cylinder before. If we hold u constant (say u=0, so x=R and y=0), then when we vary v, you trace out a vertical line parallel to the z axis with x coordinate R and y coordinate 0.

 

[Cyrus]

Yes, I see what you mean now Galileo So we take the partial derivative of the vector function [tex] \vec r (u,v) [/tex].

 

[Cyrus]

Ok, a few questions:

First, when you take the partial derivative of a scalar function, you get a scalar value, and this value is the SLOPE of the tangent to the grid curve at that point.

Second, when you take the partial derivative of a parametric function, you get a vector in return. But this vector is the tangent vector.

But I thought the Definition of a partial derivative means that you must get a scalar value, because the derivative is a slope, but we got a vector. So how does it make sense to talk about the partial derivative of a parametric function of two variables.

 

[FulhamFan3]

Ok, a few questions:

First, when you take the partial derivative of a scalar function, you get a scalar value, and this value is the SLOPE of the tangent to the grid curve at that point.

Second, when you take the partial derivative of a parametric function, you get a vector in return. But this vector is the tangent vector.

But I thought the Definition of a partial derivative means that you must get a scalar value, because the derivative is a slope, but we got a vector. So how does it make sense to talk about the partial derivative of a parametric function of two variables.

That's the definition if you take the partial of a scalar function. If you take the derivative of a vector you get a vector. It makes sense if you want to study the equation by isolating one variable.

 

[Cyrus]

I see, but here is my only concern:

[tex]\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)[/tex]

You see, if this were a scalar function, then it would mean, we approximate the value of the tangent line, which is the slope as x or y varies while the other is held fixed.

But now look at the parametric function, its clear that the partial derivative of u or v, on the surface is not a straight line, as galileo pointed out, it could be a curve. So the tangent now has components in x,y,and/or z. So its no longer meaningful to talk about the slope anymore. Is the components of the partial derivative the slope in the i,j and k directions?

 

[Galileo]

Per definition:
[tex]\vec r_u(u,v)=\lim_{\Delta u \to 0}\frac{\vec r(u+\Delta u,v)-\vec r(u,v)}{\Delta u}[/tex]
It tells you how fast the vector [itex]\vec r[/itex] is changing in the point [itex]r(u,v)[/itex] as you vary u. It's very similar to the vector function in 1D, where [itex]\frac{d}{dt}\vec r(t)[/itex] gives you the speed at which [itex]\vec r[/itex] traces out the curve at the point [itex]\vec r(t)[/itex].
So the derivatives of the vectors do NOT give the slope of the surface.

 

[Cyrus]

Hmm, let's see if this is right or not.

In the case of x=x, y=y and z=f(x,y), when you do the partial derivative you get a scalar, (the slope).

When you do r(u,v) you get a vector.

Is the "slope" of the scalar case, equal to the magnitude of the vector in the r(u,v) case? I hope this is clear, is the scalar a special case of the parametric form, (since it has only two components to the tangent, for example a change in x and z, or y and z.).

Hmm, let me state it like this. Let's say I wrote a function as r(x,y)= xi+yj+f(x,y)k.
Now there's two ways I can look at this right? If I just do the partial derivatve of [tex] f_x (x,y) [/tex] I will get some scalar value that is the slope, right? But let's say I also do the partial derivative of this vector [tex] r_x(x,y) [/tex] Now I DONT get a scalar slope, but a vector tangent, right? But what if I found the MAGNITUDE of this tangent, would I get the same anwser as the scalar slope of [tex] f_x(x,y) [/tex]? I hope this example made my question clear for you.

Edit: after looking it over I don't think what I said is correct; however, can it be interpreted in this way, when i do [tex]f_x(x,y) [/tex] I get the SLOPE of the tangent vector, but I don't know what the vector is, where it points, or what is magnitude is. When I do [tex] r_x (x,y) [/tex] I know the tangent vector at that point, I know its components in the i,j and k directions, and I know the magnitude of the tangent vector, but I know nothing of its slope.

 

[Galileo]

I see what you mean.
Indeed, in the case where a surface if given as the graph of a function with z=f(x,y), then we can parametrize the surface with parametric equations x=x, y=y and z=f(x,y).
This is a special case of the former. In this case the domain D is simply in the xy-plane and it is the projection of the surface onto the xy-plane.
The vector [itex]\vec r(x,y)[/itex] does give an indication of [itex]f_x(x,y)[/itex], because:

[tex]\vec r(x,y)=x\vec i +y\vec j+f(x,y)\vec k[/tex]

[tex]\vec r_x(x,y)=\vec i +\frac{\partial f(x,y)}{\partial x}\vec k[/itex].

So [itex]f_x[/itex] is equal to the z-component of [itex]r_x[/itex]. (Not the magnitude of [itex]r_x[/itex], since clearly: [itex]|r_x|=\sqrt{1+f_x^2}[/itex].

Similarly:
[tex]\vec r_y(x,y)=\vec j +\frac{\partial f(x,y)}{\partial y}\vec k[/itex].

 

[Cyrus]

Another question, the tangent is defined as [tex] \frac {r(t+h) - r(t)}{h} [/tex].
But since we are dealing with vectors, wouldent [tex] r(t+h)-r(t) [/tex] be enough to find a tangent vector. What is the reason for dividing by the h? All it does is scale the tangent vector. Was this done so that it Looks similar to the scalar definition of tangent?

 

[Galileo]

You have to take the limit h ->0 in that expression for the tangent.
Just look at some arbitrary curve and draw r(t+h)-r(t). It will generally not be tangent to the curve at all. It's an approximation to the tangent of the curve which becomes better and better if h gets closer to 0.

 

[Cyrus]

You have to take the limit h ->0 in that expression for the tangent.
Just look at some arbitrary curve and draw r(t+h)-r(t). It will generally not be tangent to the curve at all. It's an approximation to the tangent of the curve which becomes better and better if h gets closer to 0.

Actually, r(t+h)-r(t) will be close to the tangent vector, if delta h is small, even if you don't divide by delta h.

But I think the reason for the delta H is that a derivative is a change in one thing with respect to another, which is why the H in the denominator is necessary.

I thought about what it means to do a partial derivative of a vector function, so please tell me if this is correct or not.

When we take the partial derivative of a vector function component wise, we will get the derivative of the i, j and k components with respect to one of the variables, u or v.

So we can think of the derivative of a vector function as the rate of change in the i,j and k directions, with respect to one of the parameters u or v.

Using your example, velocity is the derivative at any point on the surface. If I do the partial derivative of a scalar function, I get a scalar value for the speed at that point on the surface as one of the variables varies and the other is held fixed. On, the other hand, for the parametric form, the magnitude of the tangent vector is the velocity at that point. But as you showed, the two are not equal. What happened?

 

[Galileo]

Actually, r(t+h)-r(t) will be close to the tangent vector, if delta h is small, even if you don't divide by delta h.

Yeah, but for any differentiable (and thus continuous) vector function
r(t+h)-r(t) will approach 0 when h goes to zero.
It's simply the way the derivative of a vector function is defined and it bears great similarity to the definition of the derivative of a real valued function.
After this definition, you can look at the geometrical significance of the derivative, then you can see it lies on a tangent line to the curve, so we also name it the tangent vector.
That's the way I think of it, and the direction of reasoning is a bit reversed. I don't think why the tangent vector should be a derivative. I start with the derivative and see it is a tangent vector.

So we can think of the derivative of a vector function as the rate of change in the i,j and k directions, with respect to one of the parameters u or v.

Correct.

Using your example, velocity is the derivative at any point on the surface. If I do the partial derivative of a scalar function, I get a scalar value for the speed at that point on the surface as one of the variables varies and the other is held fixed. On, the other hand, for the parametric form, the magnitude of the tangent vector is the velocity at that point. But as you showed, the two are not equal. What happened?

When you take the partial derivative (wrt say x) of a scalar function, you get the slope of the surface in the x-direction. It doesn't mean much to speak of the speed of a point on the surface. This case is very similar to the case of a 1D scalar function. (check section 11.3, p 769 of Stewarts).

In the case of a parametric form, it does make sense to speak of the speed at
which a curve is traced out.

The two are not equal and there's really no reason why they should be.

 

[Cyrus]

Oh, maybe there in lies my confusion, but I thought that the speed is the derivative of a function, or the slope. So I thought that [tex] f_x(x,y) [/tex] is the speed at that point with respect to x. Because isint f'(x) the speed of a body at a point x, so would'nt the same follow for [tex]f_x(x,y) [/tex]?

 

[Galileo]

Because isint f'(x) the speed of a body at a point x, so would'nt the same follow for [tex]f_x(x,y) [/tex]?

..never heard that before. I`m not even sure what is meant by that.

f'(x) gives the slope of the tangent line at point (x,f(x)) of the graph of f.
Or more commonly; simply the slope of f at x.
It isn't the 'speed' of anything.

 

[Cyrus]

In physics, if a position function is given as f(x), then the derivative of the function, f'(x) describes the velocity of the object at any point along the curve f(x).

 

[Galileo]

Yes, that's true. I did type that example in my previous response, but then deleted it to avoid confusion.
If f(t) is the position of a particle (or object) (t denotes time), then f'(t) gives the velocity of that object at time.
This is just a physical interpretation of a particular function ofcourse, and you can't speak of the speed of a point along the curve.
To interpret a function of 2 variables as a position of an object is possible, if one parameter represents time, then partial derivation wrt time gives the speed of the object along the curve f(t,y), where y is fixed.

Anyway, it's best to view a function as an abstract mathematical object and only give it a physical meaning when it is meant to represent a physical variable. Mathematical objects don't need a particular real life applicational context to work with.

 

[Cyrus]

If you look in chapter 11, I don't remember which section, maybe 11.4?, it says that the velocity vector equals the tangent vector, and the velocity is the magnitude of the velocity vector for a function r(t). The same should follow for r(u,v), so the magnitude of the tangent vector of r(u,v) should be the velocity. But this anwser differs from the scalar anwser. I am just trying to see why I am getting two different anwsers when they should be the same.

 

[Galileo]

In the scalar case, note that f(t) can only give the position of the particle if it is restricted to move in 1 dimension.
The 3d-curve can be viewed as a generalization. It has the 1d (scalar) case as a special case.

For example. If f(t) is the position, then since it moves in 1 dimnsion we can put our x-axis along the motion of the particle and parametrize it by saying f(t) is the position of the particle along the x-axis.
[tex]\vec r(t)=f(t)\vec i[/tex]

Clearly [itex]\frac{d}{dt}\vec r(t)=\vec v(t)=f'(t)\vec i[/itex].
So it's the same as a 1d case.

Also, be aware of the following distinction:
velocity is a vector and speed is a scalar. The speed is the magnitude of the velocity.

And ofcourse. Velocity is the derivative of the position vector with respect to time. If [itex]\vec r(x,y)[/itex] is given, then [itex]\vec r_x(x,y)[/itex] can be interpreted as the speed at which the curve is traced out (when y is held constant), but this is different from the physical point of view where [itex]\vec r(x,y)[/itex] represents the position of a particle. Since the differentiation is wrt x and not time, [itex]\vec r_x[/itex] is not the speed of the particle. [itex]\vec r(x,y)[/itex] depends on the particular parametrization, which is nonsense physically.

That may have been the source of confusion.

 

[Cyrus]

Im still not getting it sigh.

Im not getting consistency. Let's follow this simple example, maybe it will help.

If we have a function given by, y=f(x), then the rate of change is given by y'=f'(x).
So if the position of a particle is given by s=f(x), then the speed at any point is given by, v=y'=f'(x).

Now, let's write this as a vector function, s(x)=x(i)+y(j)+o(k) = x(i)+f(x)(j)+0k

so if we take the derivative, we get v(x)=1(i)+y'(j)+o(k) = 1(i)+f'(x)(j)+0k

Now, this means the rate of change of the position vector, in the (i) and (j) directions, in other words, its the speed in the (i) and (j) directions. So the total of these two, [tex] \sqrt{1^2 + f'(x)^2} [/tex]. But that is not the same as the other case.

Is a reason that in the scalar case, we only restrict our attention to the change in the y direction, and ignore the change in the x direction? and both the x, and y for the 3d case? ...I don't know anymore...

 

[Galileo]

Cyrus. The reason why the two results don't agree is simply because they are not describing the same case

Suppose a particle is restricted to move in 1 dimension. Then we can use the scalar form for the position of the particle, becuase that's easier.
So if x(t) gives the position of the particle at time t, then x'(t)=v(t) will give the velocity.
Note that it is sign sensitive. If x(t) decreases, then the velocity is negative.
The speed of the particle however is not x'(t), but |x'(t)|. (speed is never negative).

We could also give it in vector form:
[tex]\vec r(t)=x(t)\vec i[/tex]

the components in the y and z-directions are 0, becuase there's only motion in 1 dimension. (Along the x-axis in this case).

The velocity is: [itex]\vec r'(t)=x'(t) \vec i[/itex].
It points in the positive x-direction if x'(t) is positive and in the negative x-direction if x'(t) is negative. The speed of the particle is:
[tex]|\vec r'(t)|=\sqrt{(x'(t))^2}=|x'(t)|[/tex]
which is the same as the scalar case.

In general if the position of a particle is [itex]\vec r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k[/itex], then the speed of the particle is:
[tex]|\vec r'(t)|=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}=\sqrt{v_x^2(t)+v_y^2(t)+v_z^2(t)}[/tex]

which is an extension of the 1D case.

 

[cepheid]

Yeah Cyrus,

your first case just plots position s as a function of time, giving you a nice graph. But that curve doesn't have anything to do with the particle's actual path in space. y is a dependent variable. That's all.

In the second case, you made it so that the pariticle actually *exists* in R^2 and was moving along a trajectory given by that first curve. A completely different scenario. y is a spatial coordinate now.

 

[Cyrus]

Thanks guys, I think the thing that was messing with my brain was the fact that the two graphs look the same physically. But also, if you look at the surface area problem in the stewart text, they define the position vector as, xi+yj+f(x,y)k. So how come its not just 0i+0j+f(x,y)k ?