What is the distance between the following polar coordinates?: # (6,pi/3), (0,pi/2) #

Question

What is the distance between the following polar coordinates?: #   (6,pi/3), (0,pi/2)   #

Scarlett Allison · Accepted Answer

6.

In cartesian coordinates #(x,y)#, the distance #(d)# between two points #(x_1,y_1)# and #(x_2,y_2)# is given by: #d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2#. Now let us transform that to polar coordinates. In this new system, we know that #x = rhocostheta# and #y = rhosintheta#. Then: #d^2 = (rho_2costheta_2 - rho_1costheta_1)^2 + (rho_2sintheta_2 -rho_1sintheta_1)^2#. With a few algebric steps, we can rewrite the above expression as: #d^2 = rho_2^2(cos^2theta_2 + sin^2theta_2) + rho_1^2(cos^2theta_1 + sin^2theta_1) - 2rho_1rho_2(costheta_1costheta_2 + sintheta_1sintheta_2)# . Now, by using the trigonometric relations, we obtain the following expression for #d#: #d = sqrt(rho_1^2 + rho_2^2 - 2rho_1rho_2cos(theta_2 - theta_1)#. Then, applying our points #(rho_1,theta_1) = (6,pi/3)# and #(rho_2,theta_2) = (0,pi/2)#: #d = sqrt(36 + 0 - 0) = 6#.

Elijah Abram · Answer

undefined To find the distance between two polar coordinates, you can use the formula:

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

where $r_1$ and $r_2$ are the radii of the polar coordinates, and $\theta_1$ and $\theta_2$ are the angles.

Given the coordinates (6, π/3) and (0, π/2):

$$r_1 = 6$$
$$r_2 = 0$$
$$\theta_1 = \frac{\pi}{3}$$
$$\theta_2 = \frac{\pi}{2}$$

Using the formula:

$$d = \sqrt{6^2 + 0^2 - 2(6)(0)\cos(\frac{\pi}{2} - \frac{\pi}{3})}$$

$$d = \sqrt{36 + 0 - 2(6)(0)\cos(\frac{\pi}{6})}$$

$$d = \sqrt{36}$$

$$d = 6$$

So, the distance between the given polar coordinates is 6 units.