Calculating Surface Area Using Double Integrals

[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.
\vec r(t)=f(t)\vec i

Clearly \frac{d}{dt}\vec r(t)=\vec v(t)=f'(t)\vec i.
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 \vec r(x,y) is given, then \vec r_x(x,y) 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 \vec r(x,y) represents the position of a particle. Since the differentiation is wrt x and not time, \vec r_x is not the speed of the particle. \vec r(x,y) 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, \sqrt{1^2 + f'(x)^2}. 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, because 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:
\vec r(t)=x(t)\vec i

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

The velocity is: \vec r'(t)=x'(t) \vec i.
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:
|\vec r'(t)|=\sqrt{(x'(t))^2}=|x'(t)|
which is the same as the scalar case.

In general if the position of a particle is \vec r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k, then the speed of the particle is:
|\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)}

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 ?

 

[Cyrus]

I see what your saying about the case for t. But what if Y is not a function of time t, but is a function of the x value. Now if we take the derivative of Y w.r.t x we get a function, y'(x).

 

[Galileo]

Then, unless x represents time (which is a rather unusual notation), you cannot interpret y'(x) as a rate of change in time.