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You were really close! I would strongly recommend plugging in $t = 1$ as soon as you find your derivatives. So our derivatives are:

$$ \vec r(t) = \langle 0, 2t, 1 \rangle $$ $$ \vec s(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} t \right), 1 \right\rangle $$

Plugging in $t = 1$:

$$ \vec r(1) = \langle 0, 2(1), 1 \rangle = \langle 0, 2, 1 \rangle $$ $$ \vec s(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} (1)\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} (1)\right), 1 \right\rangle = \left\langle -\frac{\pi}{2}, 0, 1 \right\rangle $$

Now we shall plug it in to our favorite equation involving the dot product:

$$ \cos \theta = \frac{\langle 0, 2, 1 \rangle \cdot \langle -\pi/2, 0, 1 \rangle}{\left|\left| \langle 0, 2, 1 \rangle \right|\right| \left|\left| \langle -\pi/2, 0, 1 \rangle \right|\right|} = \frac{0 * (-\pi/2) + 2 * 0 + 1 * 1 }{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

That's kinda a crazy fraction... so the book decided to be all clever and simplify. Here is how they simplified. First, multiply by $\frac{2}{2}$.

$$ \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} * \frac{2}{2} = \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

This allows us to pull the $2$ into the second radical as a $4$ (since $\sqrt{4} = 2$). I also distribute the four under the radical in this step.

$$ \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{2}{\sqrt{5} * \sqrt{4* ((\pi^2/4) + 1)}} = \frac{2}{\sqrt{5} * \sqrt{\pi^2 + 4}} $$

This is the same thing as:

$$ 2 \big/ \sqrt{5} \big / \sqrt{\pi^2 + 4} $$

Which is what the book has! You got the calculus part 100% right... sometimes just the algebra can be a pain in the neck.

You were really close! I would strongly recommend plugging in $t = 1$ as soon as you find your derivatives. So our derivatives are:

$$ \vec r(t) = \langle 0, 2t, 1 \rangle $$ $$ \vec s(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} t \right), 1 \right\rangle $$

Plugging in $t = 1$:

$$ \vec r(1) = \langle 0, 2(1), 1 \rangle = \langle 0, 2, 1 \rangle $$ $$ \vec s(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} (1)\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} (1)\right), 1 \right\rangle = \left\langle -\frac{\pi}{2}, 0, 1 \right\rangle $$

Now we shall plug it in to our favorite equation involving the dot product:

$$ \cos \theta = \frac{\langle 0, 2, 1 \rangle \cdot \langle -\pi/2, 0, 1 \rangle}{\left|\left| \langle 0, 2, 1 \rangle \right|\right| \left|\left| \langle -\pi/2, 0, 1 \rangle \right|\right|} = \frac{0 * (-\pi/2) + 2 * 0 + 1 * 1 }{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

That's kinda a crazy fraction... so the book decided to be all clever and simplify. Here is how they simplified. First, multiply by $\frac{2}{2}$.

$$ \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} * \frac{2}{2} = \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

This allows us to pull the $2$ into the second radical as a $4$ (since $\sqrt{4} = 2$). I also distribute the four under the radical in this step.

$$ \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{2}{\sqrt{5} * \sqrt{4* ((\pi^2/4) + 1)}} = \frac{2}{\sqrt{5} * \sqrt{\pi^2 + 4}} $$

This is the same thing as:

$$ 2 \big/ \sqrt{5} \big / \sqrt{\pi^2 + 4} $$

Which is what the book has! You got the calculus part 100% right... sometimes just the algebra can just be a pain in the neck.

You were really close! I would strongly recommend plugging in $t = 1$ as soon as you find your derivatives. So our derivatives are:

$$ \vec r(t) r'(t) = \langle 0, 2t, 1 \rangle $$ $$ \vec s(t) s'(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} t \right), 1 \right\rangle $$

Plugging in $t = 1$:

$$ \vec r(1) r'(1) = \langle 0, 2(1), 1 \rangle = \langle 0, 2, 1 \rangle $$ $$ \vec s(t) s'(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} (1)\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} (1)\right), 1 \right\rangle = \left\langle -\frac{\pi}{2}, 0, 1 \right\rangle $$

Now we shall plug it in to our favorite equation involving the dot product:

$$ \cos \theta = \frac{\langle 0, 2, 1 \rangle \cdot \langle -\pi/2, 0, 1 \rangle}{\left|\left| \langle 0, 2, 1 \rangle \right|\right| \left|\left| \langle -\pi/2, 0, 1 \rangle \right|\right|} = \frac{0 * (-\pi/2) + 2 * 0 + 1 * 1 }{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

That's kinda a crazy fraction... so the book decided to be all clever and simplify. Here is how they simplified. First, multiply by $\frac{2}{2}$.

$$ \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} * \frac{2}{2} = \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$

This allows us to pull the $2$ into the second radical as a $4$ (since $\sqrt{4} = 2$). I also distribute the four under the radical in this step.

$$ \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{2}{\sqrt{5} * \sqrt{4* ((\pi^2/4) + 1)}} = \frac{2}{\sqrt{5} * \sqrt{\pi^2 + 4}} $$

This is the same thing as:

$$ 2 \big/ \sqrt{5} \big / \sqrt{\pi^2 + 4} $$

Which is what the book has! You got the calculus part 100% right... sometimes the algebra can just be a pain in the neck.