Problem: Let the curve $f$ be defined by $r = e^{\theta}$.

  1. Compute the slope $\frac{dy}{dx}$ of the tangent line to $f$. Then use your result to define a function $g(x,\theta)$ that is a tangent line to $f$ for every $\theta$.
  2. Find the angle $\zeta$ between $0$ and $\frac{\pi}{2}$ where the tangent line to $f(\zeta)$ intersects the $x$-axis in the point $x = 3$.

Attempt at solution:

  1. We have $$ \frac{dy}{dx} = \frac{r \cos(\theta) + \frac{dr}{d \theta} \sin(\theta)}{ -r \sin(\theta) + \frac{dr}{d \theta}\cos(\theta)} = \bigg(\frac{dy/d\theta}{dx /d \theta} \bigg) $$ hence \begin{align*} \frac{dy}{dx} = \frac{e^{\theta} \cos(\theta) + e^{\theta} \sin(\theta) } {-e^{\theta} \sin(\theta) + e^{\theta} \cos(\theta)} = m \end{align*} Now, I defined my function $g$ as follows: $$ g(x, \theta) = m(x - \theta) + e^{\theta} $$ But I'm not sure if the last part is correct, i.e. the $e^{\theta}$.

  2. If the tangent line has to go through the point $x=3$ aswell, then we must have \begin{align*} g(3,\theta) = m(3 - \theta) + e^{\theta}. \end{align*} Then, should I try solving the equation \begin{align*} m(3- \zeta) + e^{\zeta} = 0 \end{align*} for $\zeta$ with Maple, searching for solutions around the interval $\left[0, \frac{\pi}{2}\right]$ ? If I do this, I get the numerical value $1.25$, which lies between $0$ and $\frac{\pi}{2}$. So is my reasoning correct? Help would be appreciated!

  • $\begingroup$ Observe $m$ is but $\;\tan\Bigl(\theta+\dfrac\pi4\Bigr)$. $\endgroup$ – Bernard Aug 18 '15 at 10:59

One thing that may cause difficulty with this exercise is that you have to be very clear what you are talking about at each step of the solution, but the problem is not very clearly stated to begin with.

A function $g(x,\theta)$ is not a line. Even if you write $y = g(x,\theta)$ but consider $\theta$ a variable like $x,$ you do not have the equation of a line.

To get a line, first you can treat $\theta$ as a parameter that can be chosen arbitrarily but can be fixed at one arbitrary value so that $g(x,\theta)$ is effectively a function of one variable, $h(x) = g(x,\theta)$. Of course you would get a different function $h$ if you fixed $\theta$ at a different value, so it's helpful to acknowledge this in some way, for example write $h_\theta(x) = g(x,\theta).$

Next you can fix $\theta$ at a particular value long enough for you to use the equation $$y = g(x,\theta) = h_\theta(x)$$ as the equation of a line.

You have worked out the slope of the line $y = h_\theta(x).$ Note that the factors of $e^\theta$ all cancel out but the slope is still dependent on which value of $\theta$ you choose. So if $m_\theta$ is the slope of this particular line, $$m_\theta = h'_\theta(x) = \frac{dy}{dx} = \frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)}.$$

In Cartesian coordinates, the formula for a line through a point $(x_\theta,y_\theta)$ with slope $m_\theta$ is $$ y = m_\theta (x - x_\theta) + y_\theta.$$ (I have used the subscript $\theta$ here as a reminder that both the point of tangency and the slope at that point depend on the choice of $\theta.$) You know $m_\theta,$ so you just have to fill in the correct values of $x_\theta$ and $y_\theta$ for the desired point of tangency.

Here's a hint: in general, if $(x_\theta,y_\theta)$ are the Cartesian coordinates of a point on the curve $r = e^\theta,$ in the general case $x_\theta \neq \theta$ and $y_\theta \neq e^\theta.$ So try again to find the formula of the line, resisting the temptation to simply copy $\theta$ and $e^\theta$ into the formula as if they were Cartesian coordinates of something. If you choose any angle $\theta,$ and if $r = e^\theta,$ and this gives you a point whose polar coordinates are $(r, \theta),$ what are the Cartesian coordinates of that point?

For part 2, we have to figure out how the curve named $f$ relates to a function named $f.$ It seems that the intention is that the function is $f(\theta) = e^\theta$ and that the curve consists of the points with polar coordinates $(r,\theta)$ such that $r = f(\theta).$ The "tangent to $f(\zeta)$" doesn't make any sense, since the values of $f$ are numbers, not points, but if we suppose we really want the tangent to the point with polar coordinates $(f(\zeta), \zeta)$ then you just have to substitute $\theta = \zeta$ in the formulas you have already found, and try to find what $\zeta$ can be so that the tangent line passes through the point $x = 3,$ $y = 0,$ that is, $$ m_\zeta(3 - x_\zeta) + y_\zeta = 0.$$ In other words, you had almost the right approach for part 2, but the wrong values of $x_\zeta$ and $y_\zeta.$


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