Video Transcript
Convert negative two, five to polar coordinates. Give the angle in radians and round to three significant figures throughout.
So, what we’re gonna do is have a look at how we convert our coordinates from cartesian coordinates to polar coordinates. Well, we have a couple of general rules to help us. So, if we’re gonna convert from cartesian 𝑥, 𝑦 to polar coordinates 𝑟, 𝜃, then 𝑟 is gonna be equal to the square root of 𝑥 squared plus 𝑦 squared. And 𝜃 is gonna be equal to tan to the minus one, so the inverse tan, of 𝑦 divided by 𝑥.
Well, why is this? Well, let’s have a look at this diagram I’ve drawn. So, we’ve got our 𝑥- and 𝑦-coordinates. And they are negative two and five. And if we want to find 𝑟 and 𝜃, well, 𝑟 and 𝜃 are our polar coordinates. And polar coordinates tell us how far away and at what angle something is.
So, we can see that as we’ve got a right-angled triangle, we could use the Pythagorean theorem to find 𝑟. And that’d tell us the 𝑟 squared would be equal to 𝑥 squared plus 𝑦 squared because 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared, where 𝑐 squared is the hypotenuse which is the longest side opposite the right angle.
So therefore, as we wanna find 𝑟, we can take the root of each side of the equation. And if we do that, we get 𝑟 is equal to the root of 𝑥 squared plus 𝑦 squared, which is the same as we’ve got in our general rule. And similarly, because we got a right-angled triangle, and if we wanted to find the 𝜃, so the angle, then we’d use our trig ratios.
So, if we labeled our triangle, we’d have the hypotenuse, which would be 𝑟, the opposite would be our 𝑦, and the adjacent would be our 𝑥. So therefore, if we use SOH CAH TOA, our memory aid, we can see that we’d use the tangent ratio because we’ve got the opposite and the adjacent, cause we know the 𝑥- and 𝑦-coordinates. So therefore, we get tan 𝜃 is equal to the opposite over the adjacent, so 𝑦 divided by 𝑥.
So then, if we took the inverse tan of each side of the equation, cause we want to find 𝜃, we get 𝜃 is equal to the inverse tan of 𝑦 divided by 𝑥, which is exactly what we got in our general rule. Okay, great, so we’ve shown why we have these. Now let’s use them to convert from the cartesian to the polar coordinates.
So, first of all, we’ll start with the 𝑟-coordinate, where 𝑟 is equal to root negative two squared plus five squared, which is gonna be equal to root 29, which gives 5.38516 et cetera. But if we check the question, it wants us to round to three significant figures. So, the third significant figure is the eight. So, we look at the number, or digit, after it and this is our deciding number. And because this is a five or above, we’re gonna round the eight to a nine. So therefore, our 𝑟-coordinate is gonna be 5.39 to three significant figures.
Okay, now let’s find 𝜃. So, we’re gonna get 𝜃 is equal to the inverse tan of five divided by negative two. And that’s because that’s 𝑦 divided by 𝑥. It’s important to note at this point, before you put it into your calculator, that actually your calculator needs to be working in radians. That’s because the question wants the angle given in radians.
And when we do that, we get negative 1.19028, et cetera. Again, we went to round to three significant figures. Well, if we look at the third significant figure, it’s the nine. And then, after that, the digit, or the deciding number, is a zero. So, because it’s less than five, we’ll keep the nine the same. So therefore, we’ve got a value of 𝜃 which is negative 1.19 radians, and that’s to three significant figures.
But this doesn’t really make sense because if we’re looking at an angle, we’d want it to be positive. So therefore, to get the relevant positive value of 𝜃, what we’re gonna need to do is add on 𝜋. That’s because 𝜋 is the period of our tangent graph. So that’s gonna give us 1.95 radians to three significant figures. And that’s because I added on a 𝜋-value of 3.14 because that was 3.14 rounded to three significant figures because it said to round to three significant figures throughout. So therefore, negative two, five converted to polar coordinates is 5.39, 1.95. And this is rounded to three significant figures.
