By Dawkins P.

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**Extra info for Calculus III**

**Sample text**

We of course just want the line segment that starts at P and ends at Q. We can get this by simply restricting the values of t. Notice that = r ( 0) = x1 , y1 , z1 r (1) x2 , y2 , z2 So, if we restrict t to be between zero and one we will cover the line segment and we will start and end at the correct point. aspx Calculus III As noted briefly at the beginning of this section we can also vector functions of two variables. In these cases the graphs of vector function of two variables are surfaces.

Cos ϕ z= ⇒ z − 2 cos ϕ == ρ 2 − 2 3π = 2 4 ϕ= cos −1 ⇒ As with the last parts this will be the only possible ϕ in the range allowed. Finally, let’s find θ . To do this we can use the conversion for x or y. We will use the conversion for y in this case. y 1 1 2 = = = 2 ρ sin ϕ 2 2 2 2 sin θ = ⇒ π 3π θ = or θ = 4 4 Now, we actually have more possible choices for θ but all of them will reduce down to one of the two angles above since they will just be one of these two angles with one or more complete rotations around the unit circle added on.

Of course we really only need to find r and z since θ is the same in both coordinate systems. We will be able to do all of our work by looking at the right triangle shown above in our sketch. aspx Calculus III z = ρ cos ϕ r = ρ sin ϕ and these are exactly the formulas that we were looking for. So, given a point in spherical coordinates the cylindrical coordinates of the point will be, r = ρ sin ϕ θ =θ z = ρ cos ϕ Note as well that, 2 2 r 2= + z 2 ρ 2 cos 2 ϕ + ρ 2 sin = ϕ ρ 2 ( cos 2 ϕ + sin= ϕ) ρ2 Or, 2 ρ= r2 + z2 Next, let’s find the Cartesian coordinates of the same point.

### Calculus III by Dawkins P.

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