Video: Finding Limits Involving Trigonometric Functions

Convert the polar equation π‘Ÿ = 4 cos πœƒ βˆ’ 6 sin πœƒ to the rectangular form.

02:11

Video Transcript

Convert the polar equation π‘Ÿ equals four cos πœƒ minus six sin πœƒ to the rectangular form.

Remember, we convert from polar coordinates to Cartesian coordinates or rectangular coordinates using the following formulae. π‘₯ is equal to π‘Ÿ cos πœƒ and 𝑦 is equal to π‘Ÿ sin πœƒ. And these are suitable for all values of π‘Ÿ and πœƒ. Our aim is going to be to manipulate each of our equations so that we have an equation for cos πœƒ and sin πœƒ.

Well, if we divide both sides of our first equation by π‘Ÿ, we see that cos πœƒ is equal to π‘₯ over π‘Ÿ. Similarly, dividing through by π‘Ÿ in our second equation, and we find sin πœƒ equals 𝑦 over π‘Ÿ. We can then replace cos πœƒ with π‘₯ over π‘Ÿ and sin πœƒ with 𝑦 over π‘Ÿ in our original polar equation. And we see that π‘Ÿ is equal to four times π‘₯ over π‘Ÿ minus six times 𝑦 over π‘Ÿ. This simplifies to four π‘₯ over π‘Ÿ minus six 𝑦 over π‘Ÿ.

We’re next going to multiply everything through by π‘Ÿ. And we find that π‘Ÿ squared equals four π‘₯ minus six 𝑦. Now we’re obviously not quite done. We want to convert to rectangular form. This is usually of the form 𝑦 is equal to some function of π‘₯, although we’re essentially looking for an equation with π‘₯ and 𝑦 as its only variables. So we recall the other conversion formulae that we use to convert Cartesian to polar coordinates. That’s π‘Ÿ squared equals π‘₯ squared plus 𝑦 squared. We should be able to see now that we can replace π‘Ÿ squared with π‘₯ squared plus 𝑦 squared. So π‘₯ squared plus 𝑦 squared equals four π‘₯ minus six 𝑦.

We’re almost there. You might recognise this equation. We’re going to manipulate it by completing the square. We subtract four π‘₯ from both sides and add six 𝑦. Then we’re going to complete the square for π‘₯ and 𝑦. We halve the coefficient of π‘₯ β€” that gives us negative two β€” and then subtract negative two squared. So we subtract four. Similarly, we halve the coefficient of 𝑦 to get three and then subtract three squared, which is nine. And of course, this is all equal to zero. Negative four minus nine is negative 13. So we add 13 to both sides of our equation. And in rectangular form, our equation is π‘₯ minus two all squared plus 𝑦 plus three all squared equals 13.

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