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.