A particle follows the curve $r ( t ) = ( \frac{4}{t + 1}, t^3 , t^4 + 1 )$ for time $t \ge 0$
Find the point $P ( x,y,z )$ on the curve such that the normal plane to the curve at P is parallel to the plane: $\frac{-1}{9} x + 3 y + 8 z = 0$
I have no idea where to start, all I know is $(-1/9,3,8)$ would be a line orthogonal to the plane and the normal plane to the curve at $P$.
The answer is $(4/3,8,17)$ so I'm guessing they somehow found $t=2$ and plugged that in to r(t), but can someone explain to me the logical steps to take to solve this problem.
Best Answer
The derivative of the curve at $t$ is $v=(-4/(t+1)^2,3t^2,4t^3).$ Then you need to set an undetermined scalar multiple $kv$ to the vector $(-1/9,3,8)$ normal to the plane. This gives from the second coordinate that $3kt^2=3,$ so we know $kt^2=1.$ Then from the third coordinate we get $4kt^3=8,$ which means using $kt^2=1$ that $4kt^3=4(kt^2)t=4\cdot 1 \cdot t=8, $ so $t=2$ is forced. Also since $kt^2=1$ we have $k \cdot 2^2=1$ and then $k=1/4.$ Checking at the first coordinate we have $(1/4)\cdot (-4/(9))=-1/9$ as it should be.