Question

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

Answer 1

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 `(dr)/(d theta)`, which we find by deriving our original function.

To derive an exponential function with a base other than `e`, we first rewrite the original function, multiply it by the `ln` of the base, then multiply by the derivative of the term in the exponent:

`(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 `5^theta` from both terms in the radical:

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

We can now take the square root of `5^(2theta)` and pull it out of the radical:

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

The important thing to notice here is that `sqrt(1 + ln^2(5))`

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 `e`, we rewrite the original function and then divide by the `ln` of the base:

`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 `1/ln5` since it's a constant.

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`