My textbook asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to that moment.$^*$ To compensate, the textbook switched to polar coordinates and started solving it that way, which was what I have shown in the calculations below.

$*$ For further elaboration, the authors got $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$ from $x^2 + y^2 = 2Cx$. They then said, "Unfortunately, the variables cannot be separated, so without additional techniques for solving differential equations we can go no further in this direction. However, if we use polar coordinates, the equation of the family can be written as $r = 2Ccos(\theta)$". And then they continue with their calculations, as stated below.

The solution that I am getting is different from that of the textbook. All of the (similar) previous problems that I have completed have been correct, so if there are errors in my understanding of this concept, I cannot detect them.

My Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

We need to eliminate the arbitrary constant $C$ because we don't just want the orthogonal trajectory for a single curve -- we want the orthogonal trajectories for the entire family of curves; therefore, we want $\dfrac{dr}{d\theta}$ in terms of $r$ and $\theta$.

$\dfrac{r}{2\cos(\theta)} = C$

$\therefore \dfrac{dr}{d\theta} = -2\left( \dfrac{r}{2cos(\theta)} \right) sin(\theta)$

$= \dfrac{-rsin(\theta)}{cos(\theta)}$

The orthogonal trajectories will have a slope which is the negative reciprocal of the slopes of the family of curves:

$\therefore \dfrac{-d\theta}{dr} = \dfrac{-rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{d\theta}{dr} = \dfrac{rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{dr}{d\theta} = \dfrac{cos(\theta)}{rsin(\theta)}$

$\implies dr(r) = \dfrac{cos(\theta)}{sin(\theta)} (d\theta)$

And we can now proceed with separation of variables...

Textbook's Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

After eliminating C we arrive at

$\dfrac{rd\theta}{dr} = \dfrac{-cos(\theta)}{sin(\theta)}$

as the differential equation of the given family. Accordingly,

$\dfrac{rd\theta}{dr} = \dfrac{sin(\theta)}{cos(\theta)}$

is the differential equation of the orthogonal trajectories. In this case, the variables can be separated, yielding

$\dfrac{dr}{r} = \dfrac{cos(\theta) d\theta}{sin(\theta)}$

And it then proceeds with integration ...

I'm wondering if both solutions (mine and the textbook) are correct? Or have I made an error? If I've made an error, I would appreciate it if people could please take the time to carefully explain the reasoning behind it. I have only just begun studying differential equations (chapter 1), so any explanation would have to be very elementary.

My textbook, George F. Simmons' Differential Equations with Applications and Historical Notes, asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to that moment.$^*$ To compensate, the textbook switched to polar coordinates and started solving it that way, which was what I have shown in the calculations below.

$*$ For further elaboration, the authors got $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$ from $x^2 + y^2 = 2Cx$. They then said, "Unfortunately, the variables cannot be separated, so without additional techniques for solving differential equations we can go no further in this direction. However, if we use polar coordinates, the equation of the family can be written as $r = 2Ccos(\theta)$". And then they continue with their calculations, as stated below.

The solution that I am getting is different from that of the textbook. All of the (similar) previous problems that I have completed have been correct, so if there are errors in my understanding of this concept, I cannot detect them.

My Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

We need to eliminate the arbitrary constant $C$ because we don't just want the orthogonal trajectory for a single curve -- we want the orthogonal trajectories for the entire family of curves; therefore, we want $\dfrac{dr}{d\theta}$ in terms of $r$ and $\theta$.

$\dfrac{r}{2\cos(\theta)} = C$

$\therefore \dfrac{dr}{d\theta} = -2\left( \dfrac{r}{2cos(\theta)} \right) sin(\theta)$

$= \dfrac{-rsin(\theta)}{cos(\theta)}$

The orthogonal trajectories will have a slope which is the negative reciprocal of the slopes of the family of curves:

$\therefore \dfrac{-d\theta}{dr} = \dfrac{-rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{d\theta}{dr} = \dfrac{rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{dr}{d\theta} = \dfrac{cos(\theta)}{rsin(\theta)}$

$\implies dr(r) = \dfrac{cos(\theta)}{sin(\theta)} (d\theta)$

And we can now proceed with separation of variables...

Textbook's Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

After eliminating C we arrive at

$\dfrac{rd\theta}{dr} = \dfrac{-cos(\theta)}{sin(\theta)}$

as the differential equation of the given family. Accordingly,

$\dfrac{rd\theta}{dr} = \dfrac{sin(\theta)}{cos(\theta)}$

is the differential equation of the orthogonal trajectories. In this case, the variables can be separated, yielding

$\dfrac{dr}{r} = \dfrac{cos(\theta) d\theta}{sin(\theta)}$

And it then proceeds with integration ...

I'm wondering if both solutions (mine and the textbook) are correct? Or have I made an error? If I've made an error, I would appreciate it if people could please take the time to carefully explain the reasoning behind it. I have only just begun studying differential equations (chapter 1), so any explanation would have to be very elementary.

My textbook asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to that moment. To compensate, the textbook switched to polar coordinates and started solving it that way, which was what I have shown in the calculations below.

The solution that I am getting is different from that of the textbook. All of the (similar) previous problems that I have completed have been correct, so if there are errors in my understanding of this concept, I cannot detect them.

My Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

We need to eliminate the arbitrary constant $C$ because we don't just want the orthogonal trajectory for a single curve -- we want the orthogonal trajectories for the entire family of curves; therefore, we want $\dfrac{dr}{d\theta}$ in terms of $r$ and $\theta$.

$\dfrac{r}{2\cos(\theta)} = C$

$\therefore \dfrac{dr}{d\theta} = -2\left( \dfrac{r}{2cos(\theta)} \right) sin(\theta)$

$= \dfrac{-rsin(\theta)}{cos(\theta)}$

The orthogonal trajectories will have a slope which is the negative reciprocal of the slopes of the family of curves:

$\therefore \dfrac{-d\theta}{dr} = \dfrac{-rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{d\theta}{dr} = \dfrac{rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{dr}{d\theta} = \dfrac{cos(\theta)}{rsin(\theta)}$

$\implies dr(r) = \dfrac{cos(\theta)}{sin(\theta)} (d\theta)$

And we can now proceed with separation of variables...

Textbook's Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

After eliminating C we arrive at

$\dfrac{rd\theta}{dr} = \dfrac{-cos(\theta)}{sin(\theta)}$

as the differential equation of the given family. Accordingly,

$\dfrac{rd\theta}{dr} = \dfrac{sin(\theta)}{cos(\theta)}$

is the differential equation of the orthogonal trajectories. In this case, the variables can be separated, yielding

$\dfrac{dr}{r} = \dfrac{cos(\theta) d\theta}{sin(\theta)}$

And it then proceeds with integration ...

I'm wondering if both solutions (mine and the textbook) are correct? Or have I made an error? If I've made an error, I would appreciate it if people could please take the time to carefully explain the reasoning behind it. I have only just begun studying differential equations (chapter 1), so any explanation would have to be very elementary.

My textbook asks to find the orthogonal trajectory of the family of curves $r = 2Ccos(\theta)$ where C is a parameter. The original equation of the family of curves was $x^2 + y^2 = 2Cx$, but it led to an equation that was as yet unsolvable using the methods taught by the textbook up to that moment.$^*$ To compensate, the textbook switched to polar coordinates and started solving it that way, which was what I have shown in the calculations below.

$*$ For further elaboration, the authors got $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$ from $x^2 + y^2 = 2Cx$. They then said, "Unfortunately, the variables cannot be separated, so without additional techniques for solving differential equations we can go no further in this direction. However, if we use polar coordinates, the equation of the family can be written as $r = 2Ccos(\theta)$". And then they continue with their calculations, as stated below.

The solution that I am getting is different from that of the textbook. All of the (similar) previous problems that I have completed have been correct, so if there are errors in my understanding of this concept, I cannot detect them.

My Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

We need to eliminate the arbitrary constant $C$ because we don't just want the orthogonal trajectory for a single curve -- we want the orthogonal trajectories for the entire family of curves; therefore, we want $\dfrac{dr}{d\theta}$ in terms of $r$ and $\theta$.

$\dfrac{r}{2\cos(\theta)} = C$

$\therefore \dfrac{dr}{d\theta} = -2\left( \dfrac{r}{2cos(\theta)} \right) sin(\theta)$

$= \dfrac{-rsin(\theta)}{cos(\theta)}$

The orthogonal trajectories will have a slope which is the negative reciprocal of the slopes of the family of curves:

$\therefore \dfrac{-d\theta}{dr} = \dfrac{-rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{d\theta}{dr} = \dfrac{rsin(\theta)}{cos(\theta)}$

$\implies \dfrac{dr}{d\theta} = \dfrac{cos(\theta)}{rsin(\theta)}$

$\implies dr(r) = \dfrac{cos(\theta)}{sin(\theta)} (d\theta)$

And we can now proceed with separation of variables...

Textbook's Solution

$r = 2Ccos(\theta)$

$\dfrac{dr}{d\theta} = -2Csin(\theta)$

After eliminating C we arrive at

$\dfrac{rd\theta}{dr} = \dfrac{-cos(\theta)}{sin(\theta)}$

as the differential equation of the given family. Accordingly,

$\dfrac{rd\theta}{dr} = \dfrac{sin(\theta)}{cos(\theta)}$

is the differential equation of the orthogonal trajectories. In this case, the variables can be separated, yielding

$\dfrac{dr}{r} = \dfrac{cos(\theta) d\theta}{sin(\theta)}$

And it then proceeds with integration ...

I'm wondering if both solutions (mine and the textbook) are correct? Or have I made an error? If I've made an error, I would appreciate it if people could please take the time to carefully explain the reasoning behind it. I have only just begun studying differential equations (chapter 1), so any explanation would have to be very elementary.