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.

**Find the tangent**

Find the equation of the line tangent to the function given parametrically as

\begin{align*}

x(t)&=t-\sin(t) \\

y(t)&=1-\cos(t)

\end{align*}

at the point corresponding to \(t=\frac{\pi}{2}\).

**Visualizing**

Here we are given a parametric equation and asked to find the tangent line. Before showing how to do this, let’s first look at exactly what we are being asked.

Recall that tangent line to a function at a given point is the line that goes through the given point and whose slope is equal to the derivative of \(y\) with respect to \(x\) at the point. In order to find this graphically, we would first need to graph the function. In order to graph the function, you can plot some points and connect the dots, or you can use a graphing calculator. We are just trying to get a rough representation to visualize the process, so we don’t need to be exact. We have a rough sketch below.

**Calculating the tangent line**

We have a picture of what we would like to find, the only thing left to do is to actually calculate values. In order to do so, we recall that the equation of a line is \(y=m(x-x_{0})+y_{0}\) where \(m\) is the slope and \((x_{0},y_{0})\) is any point on the line. We know that the line goes through the point \((\frac{\pi}{2}-1,1)\), so we let \(x_{0}=\frac{\pi}{2}-1\) and \(y_{0}=1\). The last thing we have to do is find the slope.

The slope of the tangent line is \(\frac{dy}{dx}\) at the given point, so we must find \(\frac{dy}{dx}\). Note that we have not been given \(y\) directly as a function of \(x\), so we can’t just take the derivative using derivative rules. However, we do have \(y\) as a function of \(t\) and \(x\) as a function of \(t\). Note that, what we can do instead, is find \(y(x)=y(x(t))\). Then, \(y\) will be a new function of \(t\) that we can find the derivative of. Using this idea and the chain rule, we can find that

\begin{align*}

\frac{d}{dt}(y(x(t)))&=\frac{dy}{dx}\frac{dx}{dt} \\

\frac{dy}{dt}&=\frac{dy}{dx}\frac{dx}{dt} \\

\frac{dy}{dx}&=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}.

\end{align*}

That is, if we want to find \(\frac{dy}{dx}\), we can just take \(\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}\). We then find that

\begin{align*}

\frac{dy}{dx}&=\frac{\sin(t)}{1-\cos(t)}, \text{ so } \\

\frac{dy}{dx}|_{t=\frac{\pi}{2}}&=1.

\end{align*}

Therefore, the equation of our tangent line will be \(y=1(x-(\frac{\pi}{2}-1))+1\). If you would like to simplify this would be \(y=x-\frac{\pi}{2}+2\).

**Conclusion**

As a final step, I would take the time to plot the line that we found as the tangent in order to compare it visually to what we got. In this case, we do see that the answer will correspond, at least up to the accuracy of our drawing.

As a note to the utility of this technique, I would point out that it was much easier to find the tangent line that it was to actually graph the function. Therefore, I we wanted to know what value the function would attain at certain points, it would be much less work to use a tangent line approximation than it would be to calculate given values of the parametric equations.

I hope you learned something and this helped you with your studying. If it did, make sure to like the post and share it with anyone else you know that may benefit from the help.

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