How do you convert the polar coordinate (-2,5pi/4) into cartesian coordinates?

Question

How do you convert the polar coordinate (-2,5pi/4) into cartesian coordinates?

2 Answers

Impact of this question

12752 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-21T11:26:27

$(- 2, \frac{5 π}{4})$ $(-2,(5pi)/4)$ [polar] $= (\sqrt{2}, \sqrt{2})$ $= (sqrt(2), sqrt(2))$ [Cartesian]

Explanation:

In Polar form:
$(- 2, \frac{5 π}{4}) = (2, \frac{π}{4})$ $(-2,(5pi)/4) = (2,pi/4)$
(draw a diagram if this isn't obvious)

For an angle of $\frac{π}{4}$ $pi/4$ the "opposite" and "adjacent" sides (i.e. $x$ $x$ and $y$ $y$ ) are equal
and since $\sqrt{x^{2} + y^{2}} =$ $sqrt(x^2+y^2) =$ the radius $= 2$ $= 2$

$\to x = y = \sqrt{2}$ $rarr x=y = sqrt(2)$

Answer 2 · 2015-10-21T11:33:47

The solution is $(- \sqrt{2}; - \sqrt{2})$ $(-sqrt(2);-sqrt(2))$ . See explanation for details.

Explanation:

The given coordinates are not correct, because $r$ $r$ cannot be negative (it is a distance between 2 points and it is either zero or a positive real number).

If the given point was $(2; \frac{5 π}{4})$ $(2;(5pi)/4)$ then corresponding Carthesian coordinates would be:

$x = r cos φ = 2 \cdot cos (\frac{5 π}{4}) = 2 \cdot (- cos (\frac{π}{4}))$ $x=rcosvarphi=2*cos((5pi)/4)=2*(-cos(pi/4))$

$= 2 \cdot (- \frac{\sqrt{2}}{2}) = - \sqrt{2}$ $=2*(-sqrt(2)/2)=-sqrt(2)$

$y = r sin φ = 2 \cdot sin (\frac{5 π}{4}) = 2 \cdot (- sin (\frac{π}{4}))$ $y=rsinvarphi=2*sin((5pi)/4)=2*(-sin(pi/4))$

$= 2 \cdot (- \frac{\sqrt{2}}{2}) = - \sqrt{2}$ $=2*(-sqrt(2)/2)=-sqrt(2)$

So the Carthesian coordinates are $(- \sqrt{2}; - \sqrt{2})$ $(-sqrt(2);-sqrt(2))$

Science

Math

Humanities

... and beyond

How do you convert the polar coordinate (-2,5pi/4) into cartesian coordinates?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question