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How do you convert #(-sqrt2, 3pi/4) # to rectangular form?

Answer 1

Oliver Atkinson

#(-sqrt2,(3pi)/4)# in rectangular coordinates is #(1,-1)#

#(r,theta)# in polar coordinates is #(rcostheta,rsintheta)# in rectangular coordinates.

Hence, #(-sqrt2,(3pi)/4)# in rectangular coordinates is

#(-sqrt2xxcos((3pi)/4),-sqrt2xxsin((3pi)/4))# or

#((-sqrt2)xx(-sqrt2/2),(-sqrt2)xxsqrt2/2)# or

#((2/2),-(2/2))# or

#(1,-1)#

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Answer 2

Caroline Adams

To convert the polar coordinates (-√2, 3π/4) to rectangular form, use the following formulas:

x = r * cos(θ) y = r * sin(θ)

where r is the radius and θ is the angle.

Given (-√2, 3π/4), r = -√2 and θ = 3π/4.

Substitute these values into the formulas:

x = (-√2) * cos(3π/4) y = (-√2) * sin(3π/4)

Calculate the values:

cos(3π/4) = √2 / 2 sin(3π/4) = √2 / 2

x = (-√2) * (√2 / 2) = -2 / 2 = -1 y = (-√2) * (√2 / 2) = -2 / 2 = -1

Therefore, the rectangular form is (-1, -1).

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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.