Question

How do you find the slope of a polar curve?

Answer 1

Hint : Here in this question, we have to find the slope of a polar curve. Slope is nothing but the tangent line. As we know that the slope for a line and it is for cartesian coordinates. By using that we have to find the slope for the polar curve.

Complete step-by-step answer :
Here we have to find the slope for the polar curve. The equation for the polar curve is defined as $r = f(\theta )$ . The slope of a line is given by $\dfrac{{dy}}{{dx}}$ . If the y is the equation of a cure or a line then by applying differentiation, we can find slope for the line. To find the slope of a polar curve. We convert the coordinates of cartesian form into polar form.
Therefore, the polar coordinates are given by $x = r\cos \theta$ and $y = r\sin \theta$
Let we consider it as equation (1) and equation (2)
$x = r\cos \theta$ -----(1)
$y = r\sin \theta$ ------(2)
Now we differentiate the equation (1) with respect to $\theta$ , we get
$\dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {r\cos \theta } \right)$
Since we have $r = f(\theta )$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {f\left( \theta \right)\cos \theta } \right)$
Here the both functions are product of $\theta$ , so we apply the product rule of differentiation we get
$\Rightarrow \dfrac{{dx}}{{d\theta }} = f\left( \theta \right)\dfrac{d}{{d\theta }}\left( {\cos \theta } \right) + \cos \theta \dfrac{d}{{d\theta }}\left( {f\left( \theta \right)} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = - f\left( \theta \right)\sin \theta + f'\left( \theta \right)\cos \theta \;$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta$ -------(3)
Now we differentiate the equation (2) with respect to $\theta$ , we get
$\dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {r\sin \theta } \right)$
Since we have $r = f(\theta )$
$\Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {f\left( \theta \right)\sin \theta } \right)$
Here the both functions are product of $\theta$ , so we apply the product rule of differentiation we get
$\Rightarrow \dfrac{{dy}}{{d\theta }} = f\left( \theta \right)\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) + \sin \theta \dfrac{d}{{d\theta }}\left( {f\left( \theta \right)} \right) \\\ \Rightarrow \dfrac{{dx}}{{d\theta }} = f\left( \theta \right)\cos \theta + f'\left( \theta \right)\sin \theta \;$
$\Rightarrow \dfrac{{dx}}{{d\theta }} = f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta$ -------(4)
Now we have to find $\dfrac{{dy}}{{dx}}$
So divide the equation (4) by equation (3) we get
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}}$
Substituting the value of equation (3) and (4) we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta }}{{f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta }}$
Hence, we can substitute $f'(\theta )$ as $\dfrac{{dr}}{{d\theta }}$
Hence, we have determined the slope of a polar curve.

Note : The cartesian coordinates are x and y. The polar coordinates are $x = r\cos \theta$ and $y = r\sin \theta$ . The slope of the line is given by $\dfrac{{dy}}{{dx}}$ . To find the slope of a polar curve. We convert the coordinates of cartesian form into polar form. The slope is a tangent line to the curve.