# How do you find the length of the polar curve #r=5^theta# ?

##### 1 Answer

You can find the length of this polar curve by applying the formula for Arc Length for Parametric Equations:

#L_ = int_a^b sqrt(r^2 + ((dr)/(d theta))^2) d theta#

Giving us an answer of:

#L = (5^(theta)sqrt(1 + ln^2(5)))/ln5 |_a^b#

**Process:**

The only extra component we need to find for this formula is

To derive an exponential function with a base other than

#(dr)/(d theta) = 5^(theta) * ln(5) * (1) = 5^(theta)ln5#

Plugging this into our formula, we have:

#L = int_a^b sqrt((5^(theta))^2 + (5^(theta)ln5)^2) d theta#

Distribute the exponent:

#L =int_a^b sqrt(5^(2theta) + 5^(2theta)ln^2(5) d theta#

We can now pull out a

#L =int_a^b sqrt(5^(2theta)(1 + ln^2(5)) d theta#

We can now take the square root of

#L =int_a^b 5^(theta)sqrt(1 + ln^2(5)) d theta#

The important thing to notice here is that

is actually a constant, which means it can be pulled out of the integral entirely:

#L =sqrt(1 + ln^2(5)) int_a^b 5^(theta)d theta#

Now to integrate this exponential function with a base other than

#int_a^b 5^(theta) d theta = 5^(theta) / ln5#

You can derive this result to make sure it's correct, knowing that you can pull out

We now have:

#L = sqrt(1 + ln^2(5))5^(theta) / ln5 |_a^b#

Simplifying, we arrive at our final answer:

#L = (5^(theta)sqrt(1 + ln^2(5)))/ln5 |_a^b#