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

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

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Answer 1 · 2014-08-31T01:47:02

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

$L_{=} \int_{a}^{b} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $L_ = int_a^b sqrt(r^2 + ((dr)/(d theta))^2) d theta$

Giving us an answer of:

$L = \frac{5^{θ} \sqrt{1 + {ln}^{2} (5)}}{ln 5} {∣ ∣ ∣ ∣ ∣}_{a}^{b}$ $L = (5^(theta)sqrt(1 + ln^2(5)))/ln5 |_a^b$

Process:

The only extra component we need to find for this formula is $\frac{d r}{d θ}$ $(dr)/(d theta)$ , which we find by deriving our original function.

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

$\frac{d r}{d θ} = 5^{θ} \cdot ln (5) \cdot (1) = 5^{θ} ln 5$ $(dr)/(d theta) = 5^(theta) * ln(5) * (1) = 5^(theta)ln5$

Plugging this into our formula, we have:

$L = \int_{a}^{b} \sqrt{{(5^{θ})}^{2} + {(5^{θ} ln 5)}^{2}} d θ$ $L = int_a^b sqrt((5^(theta))^2 + (5^(theta)ln5)^2) d theta$

Distribute the exponent:

$L = \int_{a}^{b} \sqrt{5^{2 θ} + 5^{2 θ} {ln}^{2} (5) d θ}$ $L =int_a^b sqrt(5^(2theta) + 5^(2theta)ln^2(5) d theta$

We can now pull out a $5^{θ}$ $5^theta$ from both terms in the radical:

$L = \int_{a}^{b} \sqrt{5^{2 θ} (1 + {ln}^{2} (5)) d θ}$ $L =int_a^b sqrt(5^(2theta)(1 + ln^2(5)) d theta$

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

$L = \int_{a}^{b} 5^{θ} \sqrt{1 + {ln}^{2} (5)} d θ$ $L =int_a^b 5^(theta)sqrt(1 + ln^2(5)) d theta$

The important thing to notice here is that $\sqrt{1 + {ln}^{2} (5)}$ $sqrt(1 + ln^2(5))$

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

$L = \sqrt{1 + {ln}^{2} (5)} \int_{a}^{b} 5^{θ} d θ$ $L =sqrt(1 + ln^2(5)) int_a^b 5^(theta)d theta$

Now to integrate this exponential function with a base other than $e$ $e$ , we rewrite the original function and then divide by the $ln$ $ln$ of the base:

$\int_{a}^{b} 5^{θ} d θ = \frac{5^{θ}}{ln 5}$ $int_a^b 5^(theta) d theta = 5^(theta) / ln5$

You can derive this result to make sure it's correct, knowing that you can pull out $\frac{1}{ln 5}$ $1/ln5$ since it's a constant.

We now have:

$L = \sqrt{1 + {ln}^{2} (5)} \frac{5^{θ}}{ln 5} {∣ ∣ ∣}_{a}^{b}$ $L = sqrt(1 + ln^2(5))5^(theta) / ln5 |_a^b$

Simplifying, we arrive at our final answer:

$L = \frac{5^{θ} \sqrt{1 + {ln}^{2} (5)}}{ln 5} {∣ ∣ ∣ ∣ ∣}_{a}^{b}$ $L = (5^(theta)sqrt(1 + ln^2(5)))/ln5 |_a^b$

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

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

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