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How do you convert #2=(3x+7y)^2-x# into polar form?

Answer 1

James Adkins

#29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0#

#2=(3x+7y)^2-x#

#9x^2+42xy+49y^2-x=2#

After using #x=rcos(theta)# and #y=rsin(theta)# transormation,

#9r^2(cos(theta))^2+42r^2*cos(theta)*sin(theta)+49r^2(sin(theta))^2-rcos(theta)=2#

#9r^2*(1+cos(2theta))/2+21r^2*sin(2theta)+49r^2(1-cos(2theta))/2-rcos(theta)=2#

#29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0#

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Answer 2

Asher Friedman

To convert the equation (2=(3x+7y)^2-x) into polar form, you can substitute (x) and (y) with their respective polar coordinate representations:

[x = r \cos(\theta)] [y = r \sin(\theta)]

Then, substitute these expressions into the given equation. After that, replace (r^2) with (x^2 + y^2), and express (r) in terms of (x) and (y). This substitution allows you to rewrite the equation entirely in terms of (r) and (\theta), which is the polar form.

Finally, simplify the equation as much as possible to ensure it's in its standard polar form.

This process will yield the equation in polar form.

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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.