How Do You Calculate the Length of a Cardioid in the First Quadrant?

[stau40]

Homework Statement

Find the length of the cardioid with equation r = 1 + cos (theta) located in the first quadrant

Homework Equations

f (theta) = 1 + cos (theta) f'(theta) = -sin (theta) s = antiderivative (0 to (pi/2)) sq rt (f(theta)^(2) + f'(theta)^(2)) d(theta)

The Attempt at a Solution

I have worked thru the problem and have arrived at sq rt (2) antiderivative (0 to (pi/2)) sq rt (1-sin (theta)) d(theta) but have hit a road block here. Is there a trig identity that will help solve this? I haven't had any luck finding anything and don't know how else to proceed. Thanks in advance!

 

[rock.freak667]

$$1-sin x \times \frac{1+sinx}{1+sinx} = ?$$

then use cos²x+sin²x = 1

then a trig substitution.

 

[stau40]

It's been a long day, but I don't understand where the (1+sin x)/(1+sin x) comes from? From what I see that equals 1 and again leaves me with 1-sin x?

 

[Mark44]

Yes, 1 + sinx over itself equals 1, and you can always multiply by 1. If you carry out the multiplication suggested by rock.freak667, what do you get?

 

[rock.freak667]

It's been a long day, but I don't understand where the (1+sin x)/(1+sin x) comes from? From what I see that equals 1 and again leaves me with 1-sin x?

multiply it by 1-sinx and then what would you get? (you are multiplying by '1' so that you are not changing the integral)

 

[stau40]

I'm getting ((1-sinx)*(1+sinx))/1+sinx = (1+sinx-sinx-sinx^(2))/(1+sinx) = (1-sinx^(2))/(1+sinx) = (cosx^(2))/(1+sinx). Is this correct, or did I mess it up?

 

[rock.freak667]

I'm getting ((1-sinx)*(1+sinx))/1+sinx = (1+sinx-sinx-sinx^(2))/(1+sinx) = (1-sinx^(2))/(1+sinx) = (cosx^(2))/(1+sinx). Is this correct, or did I mess it up?

Yes that is correct. So now instead of integrating √(1-sinx) wrt x, you can now integrate

$$\int \sqrt{\frac{cos^2x}{1+sinx}} dx$$

now remember that √(a/b) = √a/√b

So what does the new integral become?

 

[stau40]

I'm getting the antiderivative of (cosx)/(1+(sq rt sinx))

 

[Bohrok]

For your denominator,
$$\sqrt{1 + \sin x} \neq 1 + \sqrt{\sin x}$$

 

[stau40]

How about this: (cosx)/(square root cosx)

 

[Mark44]

No, not even close. Following rock.freak667's suggestion you should have
$$\int {\frac{cos~x}{\sqrt{1+sin~x}}} dx$$

The reason for his suggestion was to get to an integrand that is relatively easy to antidifferentiate.