Calculus with Vector Functions — Question 6

Question 6 – Solution

Strategy. Square the norm to avoid differentiating a square root, then use the dot-product rule.

Step 1: Differentiate the invariant The hypothesis gives $$\mathbf r(t)\cdot\mathbf r(t)=\|\mathbf r(t)\|^2=49.$$ Differentiating both sides, $$\mathbf r'\cdot\mathbf r+\mathbf r\cdot\mathbf r'=0.$$ Because the dot product is commutative, the two terms are equal. Hence $$2\mathbf r(t)\cdot\mathbf r'(t)=0 \quad\Longrightarrow\quad \boxed{\mathbf r(t)\cdot\mathbf r'(t)=0}.$$

Step 2: Geometry The position vector points radially from the origin. When $\mathbf r'(t)\ne\mathbf 0$, it points along the tangent to the curve. Their zero dot product means the tangent is perpendicular to the radius, as expected for motion constrained to a sphere.

Step 3: Converse Now assume $\mathbf r\cdot\mathbf r'=0$. Then $$\frac d{dt}\|\mathbf r(t)\|^2 =\frac d{dt}(\mathbf r\cdot\mathbf r)=2\mathbf r\cdot\mathbf r'=0.$$ Therefore $\|\mathbf r(t)\|^2$ is constant on the interval, so its nonnegative square root $\|\mathbf r(t)\|$ is also constant. The constant is $7$ if $\|\mathbf r(t_0)\|=7$ at one point of the interval.