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b. Showing that the angle between \(\mathbf{r}\) and \(\mathbf{r}'\) varies for all \(t\). #tag_title# Step 1: Calculate the Derivative of the Vector Function#tag_content#We are given the vector function \(\mathbf{r}(t)=\langle a \cos(t), b \sin(t), c t\rangle\). To find its derivative, we differentiate each component with respect to time t: $$\mathbf{r}'(t) = \frac{d}{dt}\langle a\cos(t), b\sin(t), ct\rangle = \langle -a\sin(t), b\cos(t), c\rangle$$ #tag_title# Step 2: Find the angle between \(\mathbf{r}\) and \(\mathbf{r}'\)#tag_content#To find the angle between \(\mathbf{r}\) and \(\mathbf{r}'\), we can use the formula: $$\cos(\theta) = \frac{\mathbf{r}(t) \cdot \mathbf{r}'(t)}{||\mathbf{r}(t)|| \cdot ||\mathbf{r}'(t)||}$$ First, take the dot product of the two vectors: $$\mathbf{r}(t) \cdot \mathbf{r}'(t) = (a\cos(t)) (-a\sin(t)) + (b\sin(t)) (b\cos(t)) + (ct)c = -a^2\cos(t)\sin(t) + b^2\sin(t)\cos(t) + c^2t$$ Now, calculate the magnitudes of both vectors: $$||\mathbf{r}(t)|| = \sqrt{(a\cos(t))^2 + (b\sin(t))^2 + (ct)^2} = \sqrt{a^2\cos^2(t) + b^2\sin^2(t) + c^2t^2}$$ $$||\mathbf{r}'(t)|| = \sqrt{(-a\sin(t))^2 + (b\cos(t))^2 + c^2} = \sqrt{a^2\sin^2(t) + b^2\cos^2(t) + c^2}$$ Plug these values back into the angle formula: $$\cos(\theta) = \frac{-a^2\cos(t)\sin(t) + b^2\sin(t)\cos(t) + c^2t}{\sqrt{a^2\cos^2(t) + b^2\sin^2(t) + c^2t^2} \cdot \sqrt{a^2\sin^2(t) + b^2\cos^2(t) + c^2}}$$ Since the numerator of the expression for \(\cos(\theta)\) depends on t and contains trigonometric functions, it is not constant, implying that the angle between \(\mathbf{r}\) and \(\mathbf{r}'\) varies for all \(t\). c. Geometric interpretation of the results in parts a and b. In part a, we have shown that the angle between the vector function \(\mathbf{r}(t)=\langle at, bt, ct\rangle\) and its derivative is constant for all \(t>0\). Geometrically, this means that the vector function and its tangent (the derivative) at any point in space maintain the same angle, regardless of time. In part b, we have shown that the angle between the vector function \(\mathbf{r}(t)=\langle a\cos(t), b\sin(t), ct\rangle\) and its derivative varies for all \(t\). Geometrically, this means that as the vector function changes with time, the angle between the vector function and its tangent (the derivative) also changes. This indicates a more dynamic and potentially curved path for the vector function in comparison to part a.