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Corrected 2 mistakes (cos to sin); added a final explanation on the sign.
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Let the pivot be the point $(0,0)$ , and let's show the coordinates of the other end of the rod with $x$ and $y$. Note that $$x=Lcos\theta \quad,\quad y=Lsin\theta$$ and therefore $$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$
In time $dt$ , the rod and the wedge move as shown above. Solid lines and dashed lines correspond to beginning and end of this time interval, respectively. Suppose the sliding end of the rod moves in this time interval from $(x_1,y_1)$ to $(x_2,y_2)$ . The difference between the $x$ coordinates of these points is $dx$ , and the difference between their $y$ coordinates is $dy$ .

The horizontal displacement of the wedge in time $dt$ is $vdt$. Let the oblique edge of the wedge before and after this displacement be represented by $L_1$ and $L_2$, respectively. The equations of these two lines are: $$L_1 : \quad x = y cot \phi + b $$ $$L_2 : \quad x = y cot \phi + b + vdt $$ Note that $(x_1,y_1)$ is on $L_1$ and $(x_2,y_2)$ is on $L_2$. So: $$b = x_1 - y_1 cot \phi $$ $$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$ By rearranging and noting that $x_2 - x_1 = dx$ and $y_2 - y_1 = dy$ we have: $$dx - cot \phi dy = vdt $$ And by substitution from (1) we have: $$-L(cos\theta+cot\phi cos\theta)d\theta = vdt $$$$-L(sin\theta+cot\phi cos\theta)d\theta = vdt $$ Or: $$\frac{d\theta}{dt} = \frac{-v}{L(cos\theta+cot\phi cos\theta)}$$$$\frac{d\theta}{dt} = \frac{-v}{L(sin\theta+cot\phi cos\theta)}$$ $$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$ Finally, note that in this particular example, where the wedge is moving towards left, $v$ is negative. Therefore $\frac{d\theta}{dt}$ is positive.

enter image description here

Let the pivot be the point $(0,0)$ , and let's show the coordinates of the other end of the rod with $x$ and $y$. Note that $$x=Lcos\theta \quad,\quad y=Lsin\theta$$ and therefore $$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$
In time $dt$ , the rod and the wedge move as shown above. Solid lines and dashed lines correspond to beginning and end of this time interval, respectively. Suppose the sliding end of the rod moves in this time interval from $(x_1,y_1)$ to $(x_2,y_2)$ . The difference between the $x$ coordinates of these points is $dx$ , and the difference between their $y$ coordinates is $dy$ .

The horizontal displacement of the wedge in time $dt$ is $vdt$. Let the oblique edge of the wedge before and after this displacement be represented by $L_1$ and $L_2$, respectively. The equations of these two lines are: $$L_1 : \quad x = y cot \phi + b $$ $$L_2 : \quad x = y cot \phi + b + vdt $$ Note that $(x_1,y_1)$ is on $L_1$ and $(x_2,y_2)$ is on $L_2$. So: $$b = x_1 - y_1 cot \phi $$ $$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$ By rearranging and noting that $x_2 - x_1 = dx$ and $y_2 - y_1 = dy$ we have: $$dx - cot \phi dy = vdt $$ And by substitution from (1) we have: $$-L(cos\theta+cot\phi cos\theta)d\theta = vdt $$ Or: $$\frac{d\theta}{dt} = \frac{-v}{L(cos\theta+cot\phi cos\theta)}$$ $$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$

enter image description here

Let the pivot be the point $(0,0)$ , and let's show the coordinates of the other end of the rod with $x$ and $y$. Note that $$x=Lcos\theta \quad,\quad y=Lsin\theta$$ and therefore $$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$
In time $dt$ , the rod and the wedge move as shown above. Solid lines and dashed lines correspond to beginning and end of this time interval, respectively. Suppose the sliding end of the rod moves in this time interval from $(x_1,y_1)$ to $(x_2,y_2)$ . The difference between the $x$ coordinates of these points is $dx$ , and the difference between their $y$ coordinates is $dy$ .

The horizontal displacement of the wedge in time $dt$ is $vdt$. Let the oblique edge of the wedge before and after this displacement be represented by $L_1$ and $L_2$, respectively. The equations of these two lines are: $$L_1 : \quad x = y cot \phi + b $$ $$L_2 : \quad x = y cot \phi + b + vdt $$ Note that $(x_1,y_1)$ is on $L_1$ and $(x_2,y_2)$ is on $L_2$. So: $$b = x_1 - y_1 cot \phi $$ $$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$ By rearranging and noting that $x_2 - x_1 = dx$ and $y_2 - y_1 = dy$ we have: $$dx - cot \phi dy = vdt $$ And by substitution from (1) we have: $$-L(sin\theta+cot\phi cos\theta)d\theta = vdt $$ Or: $$\frac{d\theta}{dt} = \frac{-v}{L(sin\theta+cot\phi cos\theta)}$$ $$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$ Finally, note that in this particular example, where the wedge is moving towards left, $v$ is negative. Therefore $\frac{d\theta}{dt}$ is positive.

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Supposeenter image description here

Let the pivot isbe the point $(0,0)$ . The, and let's show the coordinates of the other end of the rod touches the wedgewith $x$ and $y$. Let's consider the image ofNote that end on$$x=Lcos\theta \quad,\quad y=Lsin\theta$$ and therefore $$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$
In time $dt$ , the horizontal axisrod and call it $x$the wedge move as shown above. Also let's use $v$Solid lines and dashed lines correspond to showbeginning and end of this time interval, respectively. Suppose the valuesliding end of the vectorrod moves in this time interval from v$(x_1,y_1)$ to $(x_2,y_2)$ . Note that inThe difference between the above figure$x$ coordinates of these points is $dx$ , and the difference between their $v$$y$ coordinates is negative (why?)$dy$ .

Given: $$v = \frac{dx}{dt}$$ We want to find: $$\frac{d \theta}{dt}$$ HereThe horizontal displacement of the wedge in time $dt$ is a way to solve$vdt$. Let the problemoblique edge of the wedge before and after this displacement be represented by $L_1$ and $L_2$, respectively. The equations of these two lines are: $$\frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt} = \frac{d \theta}{dx} \times v$$$$L_1 : \quad x = y cot \phi + b $$ So we must now find$$L_2 : \quad x = y cot \phi + b + vdt $$ Note that $\frac{d \theta}{dx}$$(x_1,y_1)$ is on ,$L_1$ and the problem stipulates, as a function of $\theta$$(x_2,y_2)$ is on $L_2$. Here is howSo: $$x = L cos \theta$$ $$dx = -L sin \theta d \theta $$$$b = x_1 - y_1 cot \phi $$ $$\frac{d\theta}{dx} = \frac{-1}{L sin \theta}$$$$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$ SoBy rearranging and noting that $x_2 - x_1 = dx$ and $y_2 - y_1 = dy$ we have: $$\frac{d \theta}{dt} = \frac{-v}{L sin \theta}$$$$dx - cot \phi dy = vdt $$ Again, note that in the given figure for the problem,And by substitution from $v$ is negative.(1) we have: $$-L(cos\theta+cot\phi cos\theta)d\theta = vdt $$ Or: $$\frac{d\theta}{dt} = \frac{-v}{L(cos\theta+cot\phi cos\theta)}$$ $$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$

Suppose the pivot is the point $(0,0)$ . The end of the rod touches the wedge. Let's consider the image of that end on the horizontal axis and call it $x$. Also let's use $v$ to show the value of the vector v . Note that in the above figure, $v$ is negative (why?).

Given: $$v = \frac{dx}{dt}$$ We want to find: $$\frac{d \theta}{dt}$$ Here is a way to solve the problem: $$\frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt} = \frac{d \theta}{dx} \times v$$ So we must now find $\frac{d \theta}{dx}$ , and the problem stipulates, as a function of $\theta$ . Here is how: $$x = L cos \theta$$ $$dx = -L sin \theta d \theta $$ $$\frac{d\theta}{dx} = \frac{-1}{L sin \theta}$$ So: $$\frac{d \theta}{dt} = \frac{-v}{L sin \theta}$$ Again, note that in the given figure for the problem, $v$ is negative.

enter image description here

Let the pivot be the point $(0,0)$ , and let's show the coordinates of the other end of the rod with $x$ and $y$. Note that $$x=Lcos\theta \quad,\quad y=Lsin\theta$$ and therefore $$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$
In time $dt$ , the rod and the wedge move as shown above. Solid lines and dashed lines correspond to beginning and end of this time interval, respectively. Suppose the sliding end of the rod moves in this time interval from $(x_1,y_1)$ to $(x_2,y_2)$ . The difference between the $x$ coordinates of these points is $dx$ , and the difference between their $y$ coordinates is $dy$ .

The horizontal displacement of the wedge in time $dt$ is $vdt$. Let the oblique edge of the wedge before and after this displacement be represented by $L_1$ and $L_2$, respectively. The equations of these two lines are: $$L_1 : \quad x = y cot \phi + b $$ $$L_2 : \quad x = y cot \phi + b + vdt $$ Note that $(x_1,y_1)$ is on $L_1$ and $(x_2,y_2)$ is on $L_2$. So: $$b = x_1 - y_1 cot \phi $$ $$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$ By rearranging and noting that $x_2 - x_1 = dx$ and $y_2 - y_1 = dy$ we have: $$dx - cot \phi dy = vdt $$ And by substitution from (1) we have: $$-L(cos\theta+cot\phi cos\theta)d\theta = vdt $$ Or: $$\frac{d\theta}{dt} = \frac{-v}{L(cos\theta+cot\phi cos\theta)}$$ $$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$

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Suppose the pivot is the point $(0,0)$ . The end of the rod touches the wedge. Let's consider the image of that end on the horizontal axis and call it $x$. Also let's use $v$ to show the value of the vector v . Note that in the above figure, $v$ is negative (why?).

Given: $$v = \frac{dx}{dt}$$ We want to find: $$\frac{d \theta}{dt}$$ Here is a way to solve the problem: $$\frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt} = \frac{d \theta}{dx} \times v$$ So we must now find $\frac{d \theta}{dx}$ , and the problem stipulates, as a function of $\theta$ . Here is how: $$x = L cos \theta$$ $$dx = -L sin \theta d \theta $$ $$\frac{d\theta}{dx} = \frac{-1}{L sin \theta}$$ So: $$\frac{d \theta}{dt} = \frac{-v}{L sin \theta}$$ Again, note that in the given figure for the problem, $v$ is negative.