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How do I go about calculating the derivative of this curve from $\mathbb{R}$ to $\mathbb{R}^3$?

$$γ(t) = (t\cdot\cos(2t), t\cdot\sin(2t), t)$$

I have tried simply taking the derivative of the three components, and adding them, but I'm not sure if this is the correct way of doing it.

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  • $\begingroup$ Use LaTeX please. $\endgroup$ – Michael Rozenberg Jun 12 at 16:51
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    $\begingroup$ You can't define a single derivative for this function, However, if you wish the calculate the "speed" of your function (as opposed to "velocity"), you might try $s(t) = \sqrt{(\frac{dx}{dt})^2+ (\frac{dy}{dt})^2+(\frac{dz}{dt})^2}$. $\endgroup$ – Hyperion Jun 12 at 16:55
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You did one step too much. Just differentiate each component and stop there. The derivative ought to be a new function $\Bbb R\to\Bbb R^3$.

Intuitively, the derivative measures how much the value of $\gamma$ changes. The values of $\gamma$ are in $\Bbb R^3$, so changes in the value of $\gamma$ are also in $\Bbb R^3$.

More rigorously, given a function $\gamma:\Bbb R\to\Bbb R^3$, with component functions $\gamma_1,\gamma_2,\gamma_3$, its derivative is by definition $$ \gamma'(t)=\lim_{h\to\infty}\frac{\gamma(t+h)-\gamma(t)}h\\ =\lim_{h\to0}\left(\frac{\gamma_1(t+h)-\gamma_1}h, \frac{\gamma_2(t+h)-\gamma_2}h, \frac{\gamma_3(t+h)-\gamma_3}h\right)\\ =(\gamma'_1(t),\gamma_2'(t),\gamma'_3(t)) $$ If you want the speed of $\gamma$, rather than the velocity, then that's simply the length of this vector. You find this the way you usually find the length of a vector: not by adding the components, but by the Pythagorean theorem.

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As mentioned by Hyperion, for a multidimensional equation of this form there are several different types of derivative. We can visualise gamma as $\gamma=\gamma(x(t),y(t),z(t))$.

You may be referring to the total derivative, which has the definition: $$\frac{d\gamma}{dt}=\frac{\partial\gamma}{\partial x}\frac{dx}{dt}+\frac{\partial\gamma}{\partial y}\frac{dy}{dt}+\frac{\partial\gamma}{\partial z}\frac{dz}{dt}$$ since we have the function defined: $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}t\cos(2t)\\t\sin(2t)\\t\end{pmatrix}$$ we are able to show that: $$\frac{d}{dt}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\cos(2t)-2t\sin(2t)\\\sin(2t)+2t\cos(2t)\\1\end{pmatrix}$$ and this can then be substituted into the formula.

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