What is the Cartesian form of -2r^3 = -theta+cot(theta)- cos(theta) −2r3=−θ+cot(θ)−cos(θ)?

1 Answer
mason m
Mar 19, 2017

-2(x^2+y^2)^(3/2)=-tan^-1(y/x)+x/y-x/sqrt(x^2+y^2)−2(x2+y2)32=−tan−1(yx)+xy−x√x2+y2

Explanation:

The polar notation arises from the following triangle:

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This tells us that r=sqrt(x^2+y^2)r=√x2+y2. We can use the triangle with legs xx and yy and angle thetaθ to rewrite any trigonometric functions.

Here, cot(theta)=1/tan(theta)=x/ycot(θ)=1tan(θ)=xy and costheta=x/r=x/sqrt(x^2+y^2)cosθ=xr=x√x2+y2.

thetaθ can be written many ways but the simplest is to note that tan(theta)=y/xtan(θ)=yx, so theta=tan^-1(y/x)θ=tan−1(yx).

We then see the equation's transformation from polar to Cartesian form:

-2r^3=-theta+cot(theta)-cos(theta)−2r3=−θ+cot(θ)−cos(θ)

-2(x^2+y^2)^(3/2)=-tan^-1(y/x)+x/y-x/sqrt(x^2+y^2)−2(x2+y2)32=−tan−1(yx)+xy−x√x2+y2

Which is in Cartesian form.

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