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What is the distance between #(-6 , pi/2 )# and #(5, pi/12 )#?

Answer 1

Grace Ashcraft

#Delta s=6.74 " "unit#

#P_A=(-6,pi/2)#

#r_A=-6#

#(r_A)^2=36#

#theta_A=pi/2#

#P_B=(5,pi/12)#

#r_B=5#

#(r_B)^2=25#

#theta_B=pi/12#

#alpha=theta_B-theta_A=pi/12-pi/2#

#alpha=(2pi-12pi)/24=-(10pi)/24=-(5pi)/12#

#"distance between two points in the polar coordinate system is given as:"#

#Delta s=sqrt(r_A^2+r_B^2-2*r_A*r_B*cos alpha)#

#Delta s=sqrt(36+25-2*6*5*cos(-(5pi)/12))#

#Delta s=sqrt(61-60*0.26)#

#Delta s=sqrt(61-15.6)#

#Delta s=6.74 " "unit#

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Answer 2

Victoria Adebayo

To find the distance between two points in a coordinate plane, we use the distance formula: ( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ).

Given the coordinates (-6, π/2) and (5, π/12), we can plug these values into the formula:

( \sqrt{(5 - (-6))^2 + (\frac{\pi}{12} - \frac{\pi}{2})^2} )

( = \sqrt{(5 + 6)^2 + (\frac{\pi}{12} - \frac{6\pi}{12})^2} )

( = \sqrt{11^2 + (\frac{-5\pi}{12})^2} )

( = \sqrt{121 + \frac{25\pi^2}{144}} )

Thus, the distance between the points (-6, π/2) and (5, π/12) is ( \sqrt{121 + \frac{25\pi^2}{144}} ).

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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.