Parametrizing Surfaces, Surface Area, and Surface Integrals Part 2

so I want to talk a little bit more generality about some of the geometry that's going on so that we can motivate setting up some of our tangent vector with respect to U and Kant your vector with respect to V so let's think about this generally in general I'm taking inputs that have coordinates in my UV space so let's say that my input is U naught V naught that's what this point is but I can think of this as being a grid of a whole bunch of u values and a whole bunch of D values just like our normal Cartesian grid and this function this parameterization function is taking this grid of use of these and it's translating it into being some surface in three-dimensional space so under this map I go from being just a flat normal looking carpet to be what I consistently call something like a magic carpet where each of these grid lines are now translated to be these curvy wavy lines right because my surface is some curvy wavy surface up here in 3-dimensional space so my point down here I'll go ahead and highlight it in color you not be not it's mapped over here to being at some some other coordinate because as an input it'll have an output now that's X of U not be not why have you not be not and zo not and the things that I want to introduce that are helpful for us are the ideas of our unit tangents so let's say that I were to consider the partial derivative just with respect to you holding my B's exactly constant so if my knees are held constant means I'm moving along this line and I have some vector T's of U and if I think of that T sub view over in this space it's going to be exactly tangent to this curve where I'm holding my DS constant and the user allowed to change similarly I think of a vector where I hold my use constant and my these are allowed to change that's going to give me some green arrow and right at that point maybe I should label these my green arrow is where I held my U naught constant and my these are allowed to change so we're going to call that T sub B and we're calling our blue one T sub u algebraically how are these two guys to find well really it's just the partial derivative of our function with respect to you maybe this looks messy but it's not really that messy so I'm just taking the derivative then holding the constant seeing how does use change similarly if I were to hold you constant and see how my these changed I would get the partial derivative of my parameterization with respect to B why is this helpful I'll tell you I'll tell you why this is helpful the final big conclusion is the fact that if I have some crazy crisscross sea surface up here it's really helpful to be able to find a normal vector that's one of the skills that we've found before that we think is helpful and it turns out that taking the cross product of my partial derivative with respect to U and my partial derivative with respect to B will always result in a normal vector and in this case the normal vector would be sort of sticking straight out of the board so it's hard for me to you know visualize what this normal vector is doing coming straight out but we have the tu crossed with T V is going to be and our normal vector at that point in the space we can evaluate it at u naught V naught and then we'll get the Nomura vector that comes out right there you're not fee not so let's look at an example of computing these tangent vectors and our normal vector let's say that we're given a parameterization of a cylinder that's given by V of UV is equal to 2 cosine u 2 sine u V so this is giving us a cylinder that looks something like this and the 2 is going to be a coefficient out front and our X in our wise so it means that it has a radius 2 and really our V can extend on infinitely in either direction so maybe I shouldn't truncate this sphere or the cylinder but we get the idea that's a cylinder and that looks like this that has radius 2 so in this case our U is representing that angle that we're tracing out so our different view values are going to give us different angles along this sphere and I'll draw my new values in blue to indicate that there's going to be these vertical lines going around in a circle whereas our V values are corresponding exactly with our height values so if I were to put the grid lines onto this our V values look something like this so if I wanted to compute what is T sub u that's asking at some fixed point here what is the slope of my tangent line when I hold my o I'm holding my be constant so it means I'm holding my height constant and I want to know what is the angle of this vector going this direction where is my T sub V I'm holding my V constant or I'm holding my U constant now and I'd say what's my rate of change in my V Direction so holding a new constant means I'm looking at this vector right here regardless it's pretty straightforward to compute this algebraically recall that this is just going to be the partial derivative of our parameterization with respect to U which we can take partial derivatives we know that the derivative of cosine is equal to negative sine so I get negative 2 sine of U the derivative of sine is cosine so I get 2 cosine U and the derivative of V with respect to U is just going to be 0 similarly when I look at my partial derivative of my parameterization with respect to V I see that there are no bees in that component there are no bees in that component and the only D I get here is 1 let's look at what the geometry of that is saying that's saying that no matter what our rate of change in the V Direction is always 1 which makes sense that the this vector is just always going to be going straight up because this is the vector that's just going straight up with unit 1 because as I as I go up this cylinder it's not like tilting outward or tilting inward or going crazy directions it's just going straight up whereas this is telling us that if we hold our radius or our height constant and we travel around in the circle we see that our tangent vector is going to change values around circle that the height is always zero because it's not going up or down or bending at all as I travel of equal height along one of those grid lines our final step let's go ahead and compute what our normal vector is going to be our normal vector recall I'll write down our definition it's going to be the cross product of T U and T V and we remember how to do cross products it's going to be our first vector is negative sine of U naked or positive 2 cosine of U 0 and then 0 0 1 and if I look at this I get I times 2 cosine nu times 1 minus 0 so this is the vector 2 cosine u I get J times negative 2 sine u so I get negative 2 sine u I know I'm not really on the board and then I get K times 0 times 0 which is 0 so what is this telling us this is telling us that our normal vector has no height component and it's pointing straight out of the board given by 2 cosine u negative 2 sine U and that's what my black normal vector is going to point right that's straight perpendicularly out at us from that cone there from that Silla

Channel: Multivariable Calculus

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Length: 9min 44sec (584 seconds)

Published: Tue Nov 19 2013