Question

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+\sqrt{9-t^{2}} \mathbf{k} $$

Short answer

The velocity vector of the object is \(\mathbf{v}(t) = \mathbf{i} + \mathbf{j} - \frac{t}{\sqrt{9 - t^2}} \mathbf{k}\), its speed is \(\sqrt{2 + \frac{t^2}{9 - t^2}}\), and the acceleration vector is \(\mathbf{a}(t) = - \frac{9}{(9 - t^2)^{3/2}} \mathbf{k}\)

Step-by-step solution

Find the velocity vector

Firstly, velocity of the object is the first derivative of the position vector function. Obtain the velocity by differentiating \(r(t) = t \mathbf{i}+t \mathbf{j}+\sqrt{9-t^{2}} \mathbf{k}\) with respect to \(t\). The derivative of \(t\) is 1, and for \(\sqrt{9 - t^2}\), use chain rule. The velocity vector \(\mathbf{v}(t)\) is found to be: \(\mathbf{v}(t) = \mathbf{i} + \mathbf{j} - \frac{t}{\sqrt{9 - t^2}} \mathbf{k}\).

Find the speed

Speed is the magnitude of the velocity vector \(\mathbf{v}(t)\). It can be calculated as the square root of the dot product of the vector with itself: \(Speed = \sqrt{\mathbf{v} \cdot \mathbf{v}}\). Calculate the speed from the above velocity function. The magnitude of vector \(\mathbf{v}(t)\) is: \(||\mathbf{v}(t)|| = \sqrt{1^2 + 1^2 + \left(\frac{t}{\sqrt{9-t^2}}\right)^2} = \sqrt{2 + \frac{t^2}{9 - t^2}}\).

Find the acceleration vector

The acceleration is simply the derivative of the velocity vector. Following similar methods as above, differentiate the velocity function \(\mathbf{v}(t) = \mathbf{i} + \mathbf{j} - \frac{t}{\sqrt{9 - t^2}} \mathbf{k}\) to get the acceleration. For computing \(- \frac{t}{\sqrt{9 - t^2}}\)'s derivative, use the quotient rule, obtaining the end result as \(\mathbf{a}(t) = - \frac{9}{(9 - t^2)^{3/2}} \mathbf{k}\).