We apply Definition 11.4.1 to find \(\vec T(t)\text{.}\) \begin{align*} \vec T(t) \amp = \frac{1}{\norm{\vrp(t)}}\vrp(t)\\ \amp =\frac{1}{\sqrt{\big(-3\sin(t) \big)^2+\big(3\cos(t) \big)^2+ 4^2}}\la -3\sin(t) ,3\cos(t) , 4\ra\\ \amp = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra. \end{align*}

We can now easily compute \(\vec T(0)\) and \(\vec T(1)\text{:}\) \begin{equation*} \vec T(0) = \la 0,\frac35,\frac45\ra\,; \vec T(1) = \la -\frac35\sin(1) ,\frac35\cos(1) ,\frac45\ra \approx \la -0.505,0.324,0.8\ra. \end{equation*}

These are plotted in Figure 11.4.3 with their initial points at \(\vec r(0)\) and \(\vec r(1)\text{,}\) respectively. (They look rather “short” since they are only length 1.)

The unit tangent vector \(\vec T(t)\) always has a magnitude of 1, though it is sometimes easy to doubt that is true. We can help solidify this thought in our minds by computing \(\norm{\vec T(1)}\text{:}\) \begin{equation*} \norm{\vec T(1)} \approx \sqrt{(-0.505)^2+0.324^2+0.8^2} = 1.000001. \end{equation*}

We have rounded in our computation of \(\vec T(1)\text{,}\) so we don't get 1 exactly. We leave it to the reader to use the exact representation of \(\vec T(1)\) to verify it has length 1.

We find \(\vrp(t) = \la 2t-1,2t+1\ra\text{,}\) and \begin{equation*} \norm{\vrp(t)} = \sqrt{(2t-1)^2+(2t+1)^2} = \sqrt{8t^2+2}. \end{equation*}

Therefore \begin{equation*} \vec T(t) = \frac{1}{\sqrt{8t^2+2}}\la 2t-1,2t+1\ra = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra. \end{equation*}

When \(t=0\text{,}\) we have \(\vec T(0) = \la -1/\sqrt{2},1/\sqrt{2}\ra\text{;}\) when \(t=1\text{,}\) we have \(\vec T(1) = \la 1/\sqrt{10}, 3/\sqrt{10}\ra\text{.}\) We leave it to the reader to verify each of these is a unit vector. They are plotted in Figure 11.4.5

In Example 11.4.2, we found \(\vec T(t) = \la (-3/5)\sin(t) ,(3/5)\cos(t) ,4/5\ra\text{.}\) Therefore \begin{equation*} \vec T\,'(t) = \la -\frac35\cos(t) ,-\frac35\sin(t) ,0\ra \text{ and } \norm{\vec T\,'(t)} = \frac35. \end{equation*}

Thus \begin{equation*} \vec N(t) = \frac{\vec T\,'(t)}{3/5} = \la -\cos(t) ,-\sin(t) ,0\ra. \end{equation*}

We compute \(\vec T(\pi/2) = \la -3/5,0,4/5\ra\) and \(\vec N(\pi/2) = \la 0,-1,0\ra\text{.}\) These are sketched in Figure 11.4.9.

In Example 11.4.4, we found \begin{equation*} \vec T(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra. \end{equation*}

Finding \(\vec T\,'(t)\) requires two applications of the Quotient Rule: \begin{align*} T\,'(t) \amp = \la \frac{\sqrt{8t^2+2}(2)-(2t-1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2},\right.\\ \amp \left.\frac{\sqrt{8t^2+2}(2)-(2t+1)\left(\frac12(8t^2+2)^{-1/2}(16t)\right)}{8t^2+2} \ra\\ \amp = \la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra \end{align*}

This is not a unit vector; to find \(\vec N(t)\text{,}\) we need to divide \(\vec T\,'(t)\) by it's magnitude. \begin{align*} \norm{\vec T\,'(t)} \amp = \sqrt{\frac{16(2t+1)^2}{(8t^2+2)^3}+\frac{16(1-2t)^2}{(8t^2+2)^3}}\\ \amp = \sqrt{\frac{16(8t^2+2)}{(8t^2+2)^3}}\\ \amp = \frac{4}{8t^2+2}. \end{align*}

Finally, \begin{align*} \vec N(t) \amp = \frac1{4/(8t^2+2)}\la \frac{4 (2 t+1)}{\left(8 t^2+2\right)^{3/2}},\frac{4 (1-2 t)}{\left(8 t^2+2\right)^{3/2}}\ra\\ \amp = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra. \end{align*}

Using this formula for \(\vec N(t)\text{,}\) we compute the unit tangent and normal vectors for \(t=-1,0\) and 1 and sketch them in Figure 11.4.11.

The previous examples give \(\vat = \la -3\cos(t) ,-3\sin(t) ,0\ra\) and \begin{equation*} \vec T(t) = \la -\frac35\sin(t) ,\frac35\cos(t) ,\frac45\ra \text{ and } \vec N(t) = \la -\cos(t) ,-\sin(t) ,0\ra. \end{equation*}

We can find \(a_\text{T}\) and \(a_\text{N}\) directly with dot products: \begin{align*} a_\text{T} \amp = \vat\cdot \vec T(t) = \frac95\cos(t) \sin(t) -\frac95\cos(t) \sin(t) +0 = 0.\\ a_\text{N} \amp = \vat\cdot \vec N(t) = 3\cos^2(t) +3\sin^2(t) + 0 = 3. \end{align*}

Thus \(\vat = 0\vec T(t) + 3\vec N(t) = 3\vec N(t)\text{,}\) which is clearly the case.

What is the practical interpretation of these numbers? \(a_\text{T} =0\) means the object is moving at a constant speed, and hence all acceleration comes in the form of direction change.

The previous examples give \(\vat = \la 2,2\ra\) and \begin{equation*} \vec T(t) = \la \frac{2t-1}{\sqrt{8t^2+2}},\frac{2t+1}{\sqrt{8t^2+2}}\ra \text{ and } \vec N(t) = \la \frac{2t+1}{\sqrt{8t^2+2}},-\frac{2t-1}{\sqrt{8t^2+2}}\ra. \end{equation*}

While we can compute \(a_\text{N}\) using \(\vec N(t)\text{,}\) we instead demonstrate using another formula from Theorem 11.4.13. \begin{align*} a_\text{T} \amp = \vat\cdot\vec T(t) = \frac{4t-2}{\sqrt{8t^2+2}}+\frac{4t+2}{\sqrt{8t^2+2}} = \frac{8t}{\sqrt{8t^2+2}}.\\ a_\text{N} \amp = \sqrt{\norm{\vat}^2-a_\text{T} ^2} = \sqrt{8-\left(\frac{8t}{\sqrt{8t^2+2}}\right)^2}=\frac{4}{\sqrt{8t^2+2}}. \end{align*}

When \(t=2\text{,}\) \(\ds a_\text{T} = \frac{16}{\sqrt{34}}\approx 2.74\) and \(\ds a_\text{N} = \frac{4}{\sqrt{34}} \approx 0.69\text{.}\) We interpret this to mean that at \(t=2\text{,}\) the particle is acclerating mostly by increasing speed, not by changing direction. As the path near \(t=2\) is relatively straight, this should make intuitive sense. Figure 11.4.16 gives a graph of the path for reference.

Contrast this with \(t=0\text{,}\) where \(a_\text{T} = 0\) and \(a_\text{N} = 4/\sqrt{2}\approx 2.82\text{.}\) Here the particle's speed is not changing and all acceleration is in the form of direction change.

Using Key Idea 11.3.15 of Section 11.3 we form the position function of the ball: \begin{equation*} \vrt = \la \big(64\cos(30^\circ) \big)t, -16t^2+\big(64\sin(30^\circ) \big)t+240\ra, \end{equation*} which we plot in Figure 11.4.18.

From this we find \(\vvt = \la 64\cos(30^\circ) , -32t+64\sin(30^\circ) \ra\) and \(\vat = \la 0,-32\ra\text{.}\) Computing \(\vec T(t)\) is not difficult, and with some simplification we find \begin{equation*} \vec T(t) = \la \frac{\sqrt{3}}{\sqrt{t^2-2t+4}}, \frac{1-t}{\sqrt{t^2-2t+4}}\ra. \end{equation*}

With \(\vat\) as simple as it is, finding \(a_\text{T}\) is also simple: \begin{equation*} a_\text{T} = \vat\cdot \vec T(t) = \frac{32t-32}{\sqrt{t^2-2t+4}}. \end{equation*}

We choose to not find \(\vec N(t)\) and find \(a_\text{N}\) through the formula \(a_\text{N} = \sqrt{\norm{\vat}^2-a_\text{T} ^2\,}\) : \begin{equation*} a_\text{N} = \sqrt{32^2-\left(\frac{32t-32}{\sqrt{t^2-2t+4}}\right)^2} = \frac{32\sqrt{3}}{\sqrt{t^2-2t+4}}. \end{equation*}

Figure 11.4.19 gives a table of values of \(a_\text{T}\) and \(a_\text{N}\text{.}\) When \(t=0\text{,}\) we see the ball's speed is decreasing; when \(t=1\) the speed of the ball is unchanged. This corresponds to the fact that at \(t=1\) the ball reaches its highest point.

After \(t=1\) we see that \(a_\text{N}\) is decreasing in value. This is because as the ball falls, it's path becomes straighter and most of the acceleration is in the form of speeding up the ball, and not in changing its direction.