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#
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Answer 2

To find the length of the polar curve ( r = 5^\theta ), you can use the arc length formula for polar curves:

[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{{dr}}{{d\theta}}\right)^2} , d\theta ]

where ( r = 5^\theta ).

First, find ( \frac{{dr}}{{d\theta}} ):

[ \frac{{dr}}{{d\theta}} = \frac{{d}}{{d\theta}}(5^\theta) = \ln(5) \cdot 5^\theta ]

Now, substitute ( r ) and ( \frac{{dr}}{{d\theta}} ) into the arc length formula and integrate it over the appropriate range of ( \theta ) values to find the length of the curve.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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