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r times Secant theta equals -5

put into polar form

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Shawn D. | Mathtastic TutorMathtastic Tutor
5.0 5.0 (9 lesson ratings) (9)
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If you want it in polar than George and Bill are both right. But just incase you need it in RECTANGULAR then:

 

Given: rSec(θ)=-5

we know that Cos(θ)=x/r since Sec(θ) =1/Cos(θ) then Sec(θ)=r/x 

so plug rSec(θ)= r(r/x)

that means that r2/x=-5 

we know by Pythagorean theorem that r2=x2+y2

so we really have (x2+y2)/x=-5

Which is sufficient as for rectangular form. If you wanted the so solve the problem as a function of x only then it would be:

y=(-5x-x2)1/2

That is if you wanted it in RECTANGULAR FORM... but as for polar form it is already in polar form. 

Bill F. | Experienced Teacher & Tutor in Round Rock, TXExperienced Teacher & Tutor in Round Roc...
5.0 5.0 (1 lesson ratings) (1)
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Hi Jen... looks to be like this already is in polar (nor rectangular coordinate) form; just need to rearrange to solve for r:  

r * Sec(θ) = -5;  r = -5/Sec(θ) = -5 / (1/Cos(θ)) = -5Cos(θ)

George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate
5.0 5.0 (2 lesson ratings) (2)
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r sec Θ = -5

-(r/5) = cos Θ

arccos (-r/5) = Θ, Θ lies in the 2nd quadrant and depends upon the value of r.