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Calculus on Vector Functions

As we approach the end of the semester, I will be taking some time to write-up a few more post for my Calculus 2 student’s. In these posts, we will focus on functions which are not represented in the usual \(y\) is a function of \(x\) form. Instead, we will look at parametric, polar and vector representations. If you would like to view more Calculus 2 study materials, please visit them here. You can also find the YouTube videos for most of those posts here.


For this example, we will be looking at the vector valued function \(f(t)=(\sin(2t),3\cos(4t),t)\). We will then need to find both the derivative and the indefinite integral of this function.

Finding the derivative

Before we find the derivative, we first want to recall what the derivative is. That is, the definition of a derivative is the slope of the tangent line at the given point. We find this, by finding the limit of the slopes of secant lines at a given point. Using this definition, we would have that
&=\lim_{\Delta t \to 0}\frac{f(t+\Delta t)-f(t)}{\Delta t} \\
&=\lim_{\Delta t \to 0}\frac{(\sin(2(t+\Delta t))-\sin(2t),3\cos(4(t+\Delta t))-3\cos(4t),t+\Delta t-t)}{\Delta t} \\
&=\lim_{\Delta t \to 0}\left(\frac{\sin(2(t+\Delta t))-\sin(2t)}{\Delta t},\frac{3\cos(4(t+\Delta t))-3\cos(4t)}{\Delta t},\frac{t+\Delta t-t}{\Delta t}\right) \\
&=(\frac{d}{dt}\sin(2t), \frac{d}{dt}3\cos(3t), \frac{d}{dt}t).

That is, if we wish to find the derivative of a vector-valued function, we can do so by finding the vector-valued function with components being the derivatives of the components of the original. In this way, we would see that
\frac{d}{dt}f(t)&=\frac{d}{dt}(\sin(2t),3\cos(4t),t) \\
&=(\frac{d}{dt}(\sin(2t)),\frac{d}{dt}3\cos(4t),\frac{d}{dt}t) \\

When finding the derivatives, we did have to use a chain rule. If you would like to see more on the chain rule, you can do so here.

Finding the integral

Now that we know how to find the derivatives of vector-valued functions, we can also find the indefinite integrals. Recall that \(\int f(t)dt\) is precisely all functions that have a derivative \(f(t)\), which means that the indefinite integral is just the inverse operation (up to adding a constant) of differentiation. Since we were able to find derivatives component-wise, we can also do so for indefinite integrals. We, therefore, find that
\int f(t)dt&= \int (\sin(2t),3\cos(4t),t) dt\\
&=(\int \sin(2t) dt, \int 3 \cos(4t)dt, \int t dt) \\
&=\left(\frac{-1}{2}\cos(2t)+c_{1},\frac{3}{4}\sin(4t)+c_{2}, \frac{t^{2}}{2}+c_{3}\right) \\
&=\left(\frac{-1}{2}\cos(2t),\frac{3}{4}\sin(4t), \frac{t^{2}}{2}\right)+C,
where \(C\) is the constant vector \((c_{1},c_{2},c_{3})\).

If you would like to see more about how to find the anti-derivatives of each of the pieces, you can do so here.


Note that we were able to find derivatives and integrals on vector-valued functions by looking at the corresponding answers for each of the components. Seeing that we could do this followed from using the formal definitions of these, which did take some work. However, once we saw this, we were left answering very similar questions to what we had previously seen in the Calculus 2.

I hope you learned something and enjoyed the post. If you did, be sure to share it with anyone else that may need help with Calculus.

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