Question

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. If \(\mathbf{r}(t) \cdot \mathbf{r}(t)\) is a constant, then \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)=0\)

Short answer

The property \(\mathbf{r}(t)\cdot\mathbf{r}'(t)\) = 0 is valid due to the fact that the derivative of \(\mathbf{r}(t).\mathbf{r}(t)\) with respect to \(t\) is zero, applying the derivative product rule for vector-valued functions, and the commutative property of the dot product.

Step-by-step solution

Understand the given property

We are given that the dot product of vector \(\mathbf{r}(t)\) with itself is a constant. Let's denote this constant as \(k\), thus we have \(\mathbf{r}(t) \cdot \mathbf{r}(t) = k\). We are asked to prove that the dot product of vector \(\mathbf{r}(t)\) with its derivative \(\mathbf{r}'(t)\) equals zero.

Differentiate the given constant

Since \(\mathbf{r}(t) \cdot \mathbf{r}(t)=k\) is a constant, its derivative with respect to \(t\) would be zero, as the derivative of a constant value is always zero. To elaborate more, \(\frac{d}{dt} \left(\mathbf{r}(t) \cdot \mathbf{r}(t)\right) = 0\).

Use the product rule for derivatives

Using the product rule of derivatives for vector-valued functions, we have \(\frac{d}{dt} \left(\mathbf{r}(t) \cdot \mathbf{r}(t)\right) = \mathbf{r}(t) \cdot \mathbf{r}'(t) + \mathbf{r}'(t) \cdot \mathbf{r}(t)\). Considering that the dot product is commutative (meaning \(\mathbf{r}(t).\mathbf{r}'(t) = \mathbf{r}'(t).\mathbf{r}(t)\)), this simplifies to \(2\cdot\mathbf{r}(t) \cdot \mathbf{r}'(t)\)

Equate to zero and prove the property

As deduced in Step 2, the derivative of \(\mathbf{r}(t) \cdot \mathbf{r}(t)\) is zero. Thus, we have \(2\cdot\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0\). Therefore, \(\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0\). Hence, the property is fully proved.