Find the positive value for which the vector $r(t) = (8t, 3t^2,6t^2 -29)$ is perpendicular to $r'(t).$

Question

Find the positive value for which the vector $r(t) = (8t, 3t^2,6t^2 -29)$ is perpendicular to $r'(t).$

How to approach this question ?

Answer 1

We have $$r'(t) = (8,6t, 12t)$$ Two vectors are perpendicular if and only if their dot product is zero; this gives us $$r(t)\cdot r'(t) = 0$$ $$64t+18t^3+72t^3-348t=0$$ Now you just need to find a positive solution for $t$ in the above equation.

Answer 2

Condition $r(t)\cdot r'(t) = 0$ is equivalent to $\|r(t)\|^2$ stationary, i.e., with first derivative equal to 0.

$$\|r(t)\|^2=(8t)^2+(3t^2)^2+(6t^2-29)^2=45t^4 - 284t^2 + 841$$

Differentiation gives:

$$\tag{1}180 t^3 - 568 t = 4t(45t^2-142)$$

The positive root of equation $t(45 t^2-142)=0$ is the one you have:

$$t=\sqrt{\frac{142}{45}}$$

Remark: equation (1) is the same as the equation given by @florence. I just wanted to use the geometrical meaning of this condition $r(t)\cdot r'(t) = 0$, which, most often, corresponds to $\|r(t)\|^2=$ constant, for example equal to $1$, which means that the curve is the trajectory of a constant speed mobile.