Given a curve defined by :$$r(t)=(3t^2-9t,3t-3t^2,2t^2-3t+5)$$
Show that the curve is flat (it lies on a plane) and then find the equation of th eplane.
If the curve lies in a plane,then :
$$A(3t^2-9t)+B(3t-3t^2)+C(2t^2-3t+5)+D=0$$
But I don't know how to get the coefficients.
How should I continue? besides does there exist anyway to use the concept of curvature to solve the problem?
On
Take the point for $t=0$ on the plane
$$r(0) = (0,0,5)$$
Now, the position of a point on the curve with respect to this is
$$r(t) - r(0) = (3t^2 -9t, 3t-3t^2, 2t^2-3t)$$
Now this vector is alway normal to the vector $\lambda\left(-\frac{1}{6}, \frac{1}{2}, 1\right)$
Hence the equation is of a plane, with a fixed normal
On
A possible way to find the coefficients (actually, ratios of the coefficients) is to assume one of them is $1$, and write the polynomial in $$ f(t) =0 $$ form, and equate the coefficients of $1,t $and $t^2$ to be equal to $0$. This works because if $$f(t) =0,$$ then $$f(0)=f'(0) =f''(0)= 0.$$Now solve the linear equations to obtain coefficients.
On
2 proofs;
1st proof: Using the formula for the torsion:
$$\tau ={{\det \left({r',r'',r'''}\right)} \over {\left\|{r'\times r''}\right\|^{2}}},$$
the fact that vector $r'''=0$ (second degree polynomials) explains that $\tau=0 \ \iff \ $ the curve is planar.
2nd proof:
$$r(t)=\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\5\end{pmatrix}+\begin{pmatrix}-9\\ \ \ 3\\-3\end{pmatrix}t+\begin{pmatrix}3\\-3\\2\end{pmatrix}t^2$$
which is a subset of the set of
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\5\end{pmatrix}+\begin{pmatrix}-9\\ \ \ 3\\-3\end{pmatrix}u+\begin{pmatrix}3\\-3\\2\end{pmatrix}v$$
where we recognize the parametric equation of a plane.
On
Expand the equation
$$A(3t^2-9t)+B(3t-3t^2)+C(2t^2-3t+5)+D=0.$$
and show that for suitable values of the coefficients it is an identity.
On
Take the equation of the plane as $x+Ay+Bz+C=0$.(Why take $4$ variables when we only $3$ are required?)
Now the point satisfies this plane. Hence $3t^2-9t+A(3t-3t^2)+B(2t^2-3t+5)+C=0$.
Now club the coefficients of $t^2,t$ and the constant terms.
Which makes the equation $(3-3A+2B)t^2+(-9+3A-3B)t+5B+C=0$ and since this equation is true for any $t$, the coefficients of $t^2,t$ and the constant must all be $0$.
So just solve the equations $3-3A+2B=0$ , $-9+3A-3B=0$ and $5B+C=0$ to get the equation of the plane.
On
Matching $t^2$ coefficients, $3A-3B+2C=0$. Matching $t$ coefficients, $-9A+3B-3C=0$. Matching the constant, $5C+D=0$. We can solve these if we choose an overall scale, e.g. $D=1$ gives the value of $C$ which leaves us with simultaneous equations in $A,\,B$. You may prefer to rescale later for integer coefficients.
On
We have the following equations in 3d space: 3t^2-9t-x=0, -3t^2+3t-y=0, 2t^2-3t+5-z=0. Now write these equations in matrix form AT=O where A is 3×3 square matrix and T is a 3×1 matrix with elements t^2, t, 1. Now assume that det A is non zero, which means that inverse of A exists. Premultiply by inverse of A to get T=O which is a contradiction since 1 can't be equal to 0. Hence det A= 0 which gives us the required equation of the plane.
Hint: If the curve is planar, then the velocity vector and the acceleration vector must both be normal to that plane's normal vector.