Do the parametric equations x = 3t^3 + 7, y = 2 - t^3, z = 5t^3 + 1 determine a line?

Question

Can you please explain the question first? And then if you know how explain that. But I really want to figure out what the question means. Thanks.

Answer 1

Not a line necessarily but a graph, in your case you got 3 graphs that may intersect

Solve the first two ecuations and then the next two (x=y=z)

(In 3D x,y,z are the coordonates for a point)

In 2D there are two coordonates x,y; y=f(x), so a point has the coordonates A(x,f(x)); in your case B(t,x(t))

Answer 2

Yes, it represents a line in 3D space. Generally the equation of a line given three points (x,y,z) is:

(x-x1)/(x2-x1) = (y-y1)/(y2-y1) = (z-z1)/(z2-z1)

In your given problem, three equations are given. Each equation is linked to the other by the common factor or physical parameter, t. That is each equation is expressed as a function of a common parameter t. When u strip the equations of this common factor (find out t^3 in each equation and equate them), then u find that u are facing an equation of the type shown above - equation of a line in 3D. In your problem, I solved for t^3 and found out that: (x-7)/3 = (-y+2)/1 = (z-1)/5 = t^3!! This rep eqn of line in 3D!!

Ok, I believe you might be satisfied with a physical interpretation of the problem. Let us begin with a simple equation: x^2 + y^2  = c^2. This represents an equation of a circle in x-y space i.e., a graph with x-y axes and a circle centred around the origin and with radius c. the physical meaning of x-y space will depend on the units taken by x and y. For example, if x is displacement and y is time, then this circle represents a particle which is oscillating about the centre with a fixed amplitude along the X axis. The slope of the tangent to the circle at any point will give the velocity of the particle at that time. Please note that even though the trajectory of the graph is a circle, the physical motion is linear.  If the particle begins to oscillate about the y axis too, then, you will get another similar equation with t as y axis and another circle too (x1^2+y1^2 = c1^2). Now what can you do to combine both these x and y displacements into one single graph? In such a case, u make use of the common factor in both these equations, namely time 't'. For instance, if u consider the two equations i have given above, and if u subtract one from the other, you will end up with an equation like x^2-x1^2 = c3^2 which is again an equation of a circle in 2D. This final equation is only a spatial equation, not a spatio-temporal one. thus the elimination of the common parameter will give u the final equation in simple 3D space. These type of individual eqs are called parametric equations and to resolve that, u just nullify the common parameter in the individual equations and establish a relation between them.

Hope it is clear. This is as far as I can go.

Answer 3

Yes, it is the equation of a line.  This is because the parameter t is always to the third power in the parametric equations.  This might be clearer with a substitution.

Let s =  t^3

Then:

x = 3t^3 + 7

y = 2 - t^3

z = 5t^3 + 1

Becomes:

x = 3s + 7

y = 2 - s

z = 5s + 1

Which is clearly the parametric equation of a line.