Question

Find the angle \(\theta\) between \(r(t)\) and \(r^{\prime}(t)\) as a function of \(t .\) Use a graphing utility to graph \(\theta(t) .\) Use the graph to find any extrema of the function. Find any values of \(t\) at which the vectors are orthogonal. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j} $$

Short answer

The angle function \(\theta(t)\) is given by \[ \theta(t) = \arccos\left(\frac{2t^{3} + t}{\sqrt{t^{4} + t^{2}}\sqrt{4t^{2} + 1}}\right) \] The graph of this function can be found using a graphing utility which demonstrates the function's behavior over different values of t. The extrema of the function can be found by finding where the derivative of the function equals zero. To find values of t where the vectors are orthogonal (dot product equals to zero), set \(2t^{3} + t = 0\).

Step-by-step solution

Differentiate the vector

Firstly, we differentiate the function \(r(t)\) to find \(r'(t)\). We find that \[ r'(t) = 2ti + j \]

Compute the magnitudes

Next, we compute the magnitudes of \(r(t)\) and \(r'(t)\) using the formula \(\sqrt{x^{2} + y^{2}}\). The magnitude of \(r(t)\) is \(\sqrt{t^{4} + t^{2}}\) and the magnitude of \(r'(t)\) is \(\sqrt{4t^{2} + 1}\).

Compute the dot product

Then, we calculate the dot product, i.e, \(r(t)\cdot r'(t) = t^{2}(2t) + t(1) = 2t^{3} + t\)

Compute the angle

We can now compute the angle \(\theta\) between the vectors \(r(t)\) and \(r'(t)\) using the dot product formula: \(\cos(\theta) = \frac{r(t)\cdot r'(t)}{\|r(t)\|\|r'(t)\|}\), Hence, \[ \theta(t) = \arccos\left(\frac{2t^{3} + t}{\sqrt{t^{4} + t^{2}}\sqrt{4t^{2} + 1}}\right) \] This is the answer as a function of t.

Graph the function

Now graph the function θ(t) using a graphing utility. The graph will give an accurate visual of the function behaviour over the different values of t.

Find extrema and orthogonal vectors

For extrema, find values of t for which derivative of θ(t) is equal to zero. For orthogonal vectors, find \(t\) where dot product is zero i.e, \(2t^{3} + t = 0\).