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-   -   Curve equation ? (http://mymathforum.com/math/337941-curve-equation.html)

kejinatsu December 8th, 2016 09:20 AM

Curve equation ?
 
I am asked to write the equation of the curve which lies above the line y=2x and lies on the surface z=x^2+y^2

But this question seems not clear and proper. Because y=2x is a line in two dimensions and z=x^2+y^2 is the surface which is in 3 dimensional space. However, at least I should find curve equation which meets these conditions. I am really confused.

btw, I should also represent this equation as w=f(x,y,z) function because after that I am going to find normal vector and tangent vector of that curve equation at some (x,y,z) point such as (1,2,5)

skipjack December 8th, 2016 02:55 PM

(y, z) = (2x, 5x²) or (y - 2x)² + (z - 5x²)² = 0.

Country Boy December 21st, 2016 08:00 AM

Quote:

Originally Posted by kejinatsu (Post 556185)
I am asked to write the equation of the curve which lies above the line y=2x and lies on the surface z=x^2+y^2

But this question seems not clear and proper. Because y=2x is a line in two dimensions and z=x^2+y^2 is the surface which is in 3 dimensional space. However, at least I should find curve equation which meets these conditions. I am really confused.

Apparently you did not see the word "above". The line y= 2x does not lie on that surface but there are points (x, y, z)= (x, 2x, z) that do, for some values of z.
In order that (x, 2x, z) lie on $\displaystyle z= x^2+ y^2$, we must have $\displaystyle z= x^2+ (2x)^2= 5x^2$. So these points are of the form $\displaystyle (x, 2x, 5x^2)$.

Quote:

btw, I should also represent this equation as w=f(x,y,z) function because after that I am going to find normal vector and tangent vector of that curve equation at some (x,y,z) point such as (1,2,5)
No, you can't. w= f(x,y,z), since it depends on three variables is three dimensional. If you mean w as a constant, it is a surface. In either case, it is not a curve. You can write the curve in parametric form as x= t, y= 2t, z= 5t^2 or as a position vector $\displaystyle t\vec{i}+ 2t\vec{j}+ 5t^2\vec{k}$. While a smooth curve in three dimensions has a unique tangent line (so all tangent vectors are multiples of the unit tangent vectors) there is an entire plane perpendicular to the curve. Do you mean the vector normal to the surface?


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