What is the distance between the following polar coordinates?: # (5,(3pi)/4), (3,(11pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (5,(3pi)/4), (3,(11pi)/8)   #

Charles Arocha · Accepted Answer

The distance is approximately #6.74# units. See explanation.

If two polar points #A=(r_1,varphi_1)# and #B=(r_2,varphi_2)# are given, then the distance between the points can be calculated as: #d(A,B)=sqrt(r_1^2+r_2^2-2r_1r_2cos(varphi_2-varphi_1))# Here we have: #d=sqrt(5^2+3^2-2*5*3*cos((11pi)/8-(3pi)/4))=# #=sqrt(25+9-30cos((11pi-6pi)/8))=# #=sqrt(34-30cos((5pi)/8))~~sqrt(34-30*(-0.383))~~# #~~sqrt(34+11.49)~~sqrt(45.49)~~6.74#

Cameron Alexander · Answer

undefined To find the distance between two polar coordinates $ (r_1, \theta_1) $ and $ (r_2, \theta_2) $, we use the formula:

$$ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} $$

Given the polar coordinates $ (5, \frac{3\pi}{4}) $ and $ (3, \frac{11\pi}{8}) $, we substitute the values into the formula:

$$ d = \sqrt{5^2 + 3^2 - 2(5)(3)\cos\left(\frac{11\pi}{8} - \frac{3\pi}{4}\right)} $$

Solving inside the cosine function:

$$ \frac{11\pi}{8} - \frac{3\pi}{4} = \frac{22\pi - 12\pi}{16} = \frac{10\pi}{16} = \frac{5\pi}{8} $$

Substituting back into the formula:

$$ d = \sqrt{25 + 9 - 30\cos\left(\frac{5\pi}{8}\right)} $$

Now, calculate $ \cos\left(\frac{5\pi}{8}\right) $ using a calculator, then substitute it into the formula:

$$ d = \sqrt{25 + 9 - 30 \cdot \cos\left(\frac{5\pi}{8}\right)} $$

$$ d = \sqrt{25 + 9 - 30 \cdot \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right)} $$

$$ d = \sqrt{34 + 15\sqrt{2+\sqrt{2}}} $$

This is the distance between the given polar coordinates.