Question

\[x = r\cos \theta \], \[y = r\sin \theta \], then \[\dfrac{{\delta r}}{{\delta x}}\] is equal to:
A. \[\sec \theta \]
B. \[\sin \theta \]
C. \[\cos \theta \]
D. \[cosec\theta \]

Answer 1

Hint: Here, given an equation having a trigonometric function, we have to find the Partial derivative or partial differentiated term of the function. First write an equation of \[r\] by adding the x and y function and next differentiate equation \[r\] partially with respect to x by using a standard differentiation formula of trigonometric ratio and use chain rule for differentiation. And on further simplification we get the required differentiate value.

Complete step-by-step answer:
The process of finding the partial derivatives of a given function is called the partial differentiation.
Consider the given question:
\[ \Rightarrow \,\,\,x = r\cos \theta \]
Squaring on both side, we have
\[ \Rightarrow \,\,\,{x^2} = {r^2}{\cos ^2}\theta \] ------(1)
And
\[ \Rightarrow \,\,\,y = r\sin \theta \]
Squaring on both side, we have
\[ \Rightarrow \,\,\,{y^2} = {r^2}{\sin ^2}\theta \] ------(2)
Add equation (1) and (2), we get
\[ \Rightarrow \,\,\,{x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta \]
Take \[{r^2}\] as common on RHS, then
\[ \Rightarrow \,\,\,{x^2} + {y^2} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\]
By using the trigonometric identity: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], then we have
\[ \Rightarrow \,\,\,{x^2} + {y^2} = {r^2}\left( 1 \right)\]
On simplification, we get
\[ \Rightarrow \,\,\,{x^2} + {y^2} = {r^2}\]-------(3)
Now, differentiate the equation (3) partially with respect to x, then
\[ \Rightarrow \,\,\,\dfrac{\delta }{{\delta x}}\left( {{x^2} + {y^2}} \right) = \dfrac{\delta }{{\delta x}}\left( {{r^2}} \right)\]
When differentiating partially with respect to \[x\], then \[y\] term be the constant.
\[ \Rightarrow \,\,\,\dfrac{\delta }{{\delta x}}\left( {{x^2}} \right) + \dfrac{\delta }{{\delta x}}\left( {{y^2}} \right) = \dfrac{\delta }{{\delta x}}\left( {{r^2}} \right)\]
By using a standard differentiation formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], then we have
 \[ \Rightarrow \,\,\,2x + 0 = 2r\dfrac{{\delta r}}{{\delta x}}\]
\[ \Rightarrow \,\,\,2x = 2r\dfrac{{\delta r}}{{\delta x}}\]
Divide 2 on both side, we have
\[ \Rightarrow \,\,\,x = r\dfrac{{\delta r}}{{\delta x}}\]
Divide r on both side, then
\[ \Rightarrow \,\,\,\dfrac{{\delta r}}{{\delta x}} = \dfrac{x}{r}\]
Substitute the value \[x = r\cos \theta \], then
\[ \Rightarrow \,\,\,\dfrac{{\delta r}}{{\delta x}} = \dfrac{{r\cos \theta }}{r}\]
On simplification, we get
\[ \Rightarrow \,\,\,\dfrac{{\delta r}}{{\delta x}} = \cos \theta \]
Hence, it’s a required solution.

So, the correct answer is “Option C”.

Note: When solving differentiation based questions, must know the differentiation formulas for the trigonometry ratios and these differentiation formulas are standard. Remember if there is a function that has x and y as the variables, it will determine the partial derivative with respect of x and another variable (y) is constant and vice versa. But finding partial derivatives operates the variable one by one or depends on which one that is asked in the problem.