Assignment 10

Parametric Curves

I graph

x = t +1

y = 2t -1

for some appropraite range for t.

I obtained the following straight line in two dimension.

In a three dimension coordinate space, y=2x-3 is the equation of the blue plane in which the coeficient of z is equal to zero. The equation of this plane is independent of the values of z. thus for any values of x and y that satify the equation, any value of z will also work. I rewrite and plot this equation in parametric form to obtain the intersection of the plane with the xy plane. The intersection is a grey line on the diagram below.

To understand the above, I can try to find all the planes that that passes through all the parametric equation above. One of the lines is obviously y=2x-3. The other line that cuts z at 0. Combining both lines help us obtain all the planes that passes through the parametric equation. We can vary z in or der to obtain all the planes, ie. nz+y=2x-3. We get the same result frpm ny+z = 2nx -3n. Click here to see the result.

I use the same idea above to write the parametric equation of a line through (7,5) with slope of 3. We first find the equation of this line in 2 dimension. The line is y= 3x-16 and the parametric expression is

Using similiar method as above, the planes that pass through this parametric equation is desribed by y+nz=3x-16. Clck Here to see animation.

Cylinder

Next, I went on to investigate the parameterized curve that lies at the intersection of the plane z=nx and the cylinder

Solving these 2 equations yield the following parameterized curve.

z=0 is the green plane that cuts through the cylinder that is perpendicular to the x-y plane.

If we change the plane or cylinder , we could use the above approach to find the parameterized curve that lies at the intersection of the plane and cylinder. Suppose the cylinder is still

The plane is given by

The parameterized curve at the intersection will be

Click here to see animation. The shape of this intersection is always a circle.

Cone

I next look at the cone and its intersection with a given plane.

Suppose the cone is given by the following expression:

and the plane is given by

the parameterized curve at the intersection will be

Click here to see animation. The plane intersects the cone and produces 4 different conics. I found that in this example, the value of n that gives the different conics are as follows:

n = 0 gives a circle

gives an ellipse

gives a parabola

gives a hyperbola

To determine the value of n , we express our equation for the cone and plane into the generalized form

(ellipse)

(hyperbola)

(circle)

I obtain the follwing expression for the example above

From this expression, it is easy to see why n=1 is the magic number here.

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