Question

How will you find the length of the polar curve $r = {5^\theta }$ ?

Answer 1

Here as we need to find the length of the polar curve so we will use the formula for the arc length $(L)$ for parametric equations $L = \int_a^b {\sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}} } d\theta$ . We need to find $\dfrac{{dr}}{{d\theta }}$ ,which is the only extra component for this formula to be calculated and while finding the length of the polar curve find the correct limits of integration also.

Formula used:
The formula which we can apply for finding the length of polar curve is
$L = \int_a^b {\sqrt {{r^2} + {{\left( {\dfrac{{dr}}{{d\theta }}} \right)}^2}} } d\theta$

Complete step by step answer:
Firstly for deriving an exponential function with base other than $e$ we need to rewrite the original function and next we will multiply it by the $\ln$ of the base.Later on by multiplying by the derivative of the term in the exponent we will get,
$\dfrac{{dr}}{{d\theta }} = {5^\theta } \times \ln (5) \times (1) = {5^\theta }\ln 5$
Now after we will substitute this in the formula we get

\Rightarrow L = \int_a^b {\sqrt {{{({5^\theta })}^2} + {{({5^\theta }\ln 5)}^2}d\theta } } \\\ $$ Then we will pull out $${5^\theta }$$from both the terms and we will get $$L = \int_a^b {\sqrt {({5^{2\theta }})(1 + {{\ln }^2}(5))d\theta } } $$ Next we will pull out $${5^{2\theta }}$$from the radical $$L = \int_a^b {\sqrt {{5^{2\theta }}(1 + {{\ln }^2}(5))d\theta } } $$ Now we will take out the square of $${5^{2\theta }}$$and pull out of radical $$L = \int_a^b {5\theta \sqrt {1 + {{\ln }^2}(5)d\theta } } \\\ \Rightarrow L = \sqrt {1 + {{\ln }^2}(5)} \int_a^b {{5^\theta }d\theta } \\\ $$ Now for integrating exponential function with base other than $$e$$ rewrite the original function and and divide by the $$\ln $$ of the base $$\int_a^b {{5^\theta }d\theta = \dfrac{{{5^\theta }}}{{\ln 5}}} $$ Later on simplifying it we get $$L = \sqrt {1 + l{{\ln }^2}(5)} \left. {\dfrac{{{5^\theta }}}{{\ln 5}}} \right|_a^b \\\ \therefore L = \left. {\dfrac{{{5^\theta }\sqrt {1 + {{\ln }^2}(5)} }}{{\ln 5}}} \right|_a^b \\\ $$ **Hence,the length of the polar curve is $\left. {\dfrac{{{5^\theta }\sqrt {1 + {{\ln }^2}(5)} }}{{\ln 5}}} \right|_a^b$** **Note:** While solving such types of questions easily we need to have some understanding about trigonometric properties. We should keep in mind that the key for computing the length of a polar curve is to think of it as a parameterized curve with parameter $$\theta $$ and keep in mind that $$\theta $$ represents the polar angle and while solving we need to convert polar form into Cartesian structure.For simplify integral in above problem we have to use substitution and the range for the tangent function encompasses all real numbers.