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

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.