How to find the parametric equations for the curb z-1=x^2+y^2, x-y=0 such that z increases when x & y are positive?

Asked on by binges10

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nathanshields | High School Teacher | (Level 1) Associate Educator

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Notice that the equation of a circle is `x^2+y^2=r^2` .  So your first equation describes a set of circles where the radius = `sqrt(z-1)` .

Your graph is empty until `z>=1`...then you get a "stack" of circles whose radius is increasing.

If the radius were increasing at a linear rate, you'd have a cone.  But the radius is increasing ` `like a square root graph, so you get a nice vase shape, or bullet shape.  Imagine the graph of `y=sqrt(z-1)` rotated around the z-axis.

OK, now x - y = 0 means x = y.  So imagine the regular graph of y = x, but extend it out in the z direction to make a plane.  The intersection of the plane and the "bullet vase" is the curve we're after.

It's just a parabola!

If you projected it onto the x-z plane it would look like this:

Same thing for the y-z plane.  So let's just let

x(t) = t

y(t) = t

z(t) = t^2 + 1

Double-check the condition that z increases when x and y (in other words, t) are positive.  Looks good.

I know it's hard to visualize this stuff from words alone, so try to work it through with a series of graphs, and talk with your instructor!

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