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Sample text

We’ve got two and we will use P. We need a vector that is parallel to the line and since we’ve got two points we can find the vector between them. This vector will lie on the line and hence be parallel to the line. Also, let’s remember that we want to preserve the starting and ending point of the line segment so let’s construct the vector using the same “orientation”. r v = x2 - x1 , y2 - y1 , z2 - z1 Using this vector and the point P we get the following vector equation of the line. r r ( t ) = x1 , y1 , z1 + t x2 - x1 , y2 - y1 , z2 - z1 While this is the vector equation of the line, let’s rewrite the equation slightly.

This gives, f ( x, y ) = Ax + By + D To graph a plane we will generally find the intersection points with the three axes and then graph the triangle that connects those three points. This triangle will be a portion of the plane and it will give us a fairly decent idea on what the plane itself should look like. aspx Calculus III For purposes of graphing this it would probably be easier to write this as, z = 12 - 3 x - 4 y Þ 3x + 4 y + z = 12 Now, each of the intersection points with the three main coordinate axes is defined by the fact that two of the coordinates are zero.

Let’s first start with a point in spherical coordinates and ask what the cylindrical coordinates of the point are. So, we know ( r , q , j ) and what to find ( r , q , z ) . Of course we really only need to find r and z since q 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 = r cos j r = r sin j 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 = r sin j q =q z = r cos j Note as well that, r 2 + z 2 = r 2 cos 2 j + r 2 sin 2 j = r 2 ( cos 2 j + sin 2 j ) = r 2 Or, r 2 = r2 + z2 Next, let’s find the Cartesian coordinates of the same point.

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