Polar Coordinates for Point (-5, 3√7) | Solving Using Pythagorean Theorem

[imy786]

1. Homework Statement


A point has coordinates (−5, 3*square root of 7).
What is the polar coordinates of this point?

(in the form a,b)


2. Homework Equations


x= rcos theta
y=rsin theta

3. The Attempt at a Solution

Using phythagoras thrown

-5=x=rcos theta (eq 1)
3*square root 7=Y=rsin theta (eq 2)

r=-5/cos theta
r=3*square root 7 / sin theta

 

[Mark44]

1. Homework Statement


A point has coordinates (−5, 3*square root of 7).
What is the polar coordinates of this point?

(in the form a,b)


2. Homework Equations


x= rcos theta
y=rsin theta

3. The Attempt at a Solution

Using phythagoras thrown

I don't see that you have actually used the Pythagoras Theorem. Isn't r² = x² + y²?

It might be helpful to draw a right triangle, with base x, altitude y, and hypotenuse r.

-5=x=rcos theta (eq 1)
3*square root 7=Y=rsin theta (eq 2)

r=-5/cos theta
r=3*square root 7 / sin theta

 

[grzz]

Try to find the value of r by eliminating $\theta$ using sin$^{2}$$\theta$ + cos$^{2}$$\theta$ = 1.

 

[imy786]

(-5)^2=r^2cos^2 theta (eq 1)
25 = r^2cos^2
r^2 = 25 / cos^2
r^2 = 25 / (1 - sin ^2 theta)
r^2 * (1 - sin ^2 theta) = 25
(1 - sin ^2 theta) = 25 / r^2
1 - (25 / r^2) = sin ^2 theta

(3)^2* ( 7) =Y=r^2sin^2 theta (eq 2)
9*7 = r^2sin^2 theta
64 = r^2 (sin^2 theta)
r^2 = 64 / sin^2 theta
r^2* sin^2 theta = 64
sin^2 theta = 64 /r^2

1 - (25 / r^2) = 64 /r^2
r^2 - (25*r^2 / r^2) = 64
r^2 - (25) = 64
r^2 = 64-25
r^2 = 39
r = =/- (square root of 39)

Polar co-ordinates of the point?

 

[verty]

x = r cos(theta)
y = r sin(theta)

hmm, I know x and y but not r and theta. I have two unknowns. I need to eliminate r or theta. How can I do that?

if I can eliminate r, I can find theta.
if I can eliminate theta, I can find r.

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[Mark44]

You're really going at it the hard way.

(-5)^2=r^2cos^2 theta (eq 1)
25 = r^2cos^2
r^2 = 25 / cos^2
r^2 = 25 / (1 - sin ^2 theta)
r^2 * (1 - sin ^2 theta) = 25
(1 - sin ^2 theta) = 25 / r^2
1 - (25 / r^2) = sin ^2 theta

(3)^2* ( 7) =Y=r^2sin^2 theta (eq 2)
9*7 = r^2sin^2 theta

9*7 != 64.

64 = r^2 (sin^2 theta)
r^2 = 64 / sin^2 theta
r^2* sin^2 theta = 64
sin^2 theta = 64 /r^2

1 - (25 / r^2) = 64 /r^2
r^2 - (25*r^2 / r^2) = 64
r^2 - (25) = 64
r^2 = 64-25

Add 25 to both sides. Also, as noted above, 64 isn't the right number.

r^2 = 39
r = =/- (square root of 39)

Polar co-ordinates of the point?

If you redo your calculations for r, you will have one of the coordinates. As already recommended, draw a right triangle with base -5 (i.e., to the left of the origin), altitude 3*sqrt(7). You can find the hypotenuse as r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

 

[imy786]

Need answer in the form

(1 , pie/3)

 

[imy786]

You're really going at it the hard way.9*7 != 64.Add 25 to both sides. Also, as noted above, 64 isn't the right number.
If you redo your calculations for r, you will have one of the coordinates. As already recommended, draw a right triangle with base -5 (i.e., to the left of the origin), altitude 3*sqrt(7). You can find the hypotenuse as r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

r = $\sqrt{(25 + (21}$

r = $\sqrt{(41}$

r = 6.40

But still need polar point in the form (f, g/pie)

 

[Mark44]

r = $\sqrt{(-5)^2 + (3\sqrt{7})^2}$

r = $\sqrt{(25 + (21}$

$(3 \sqrt{7})^2 \neq 21$
Your continuing algebra errors will make completing this problem very difficult.

r = $\sqrt{(41}$

r = 6.40

But still need polar point in the form (f, g/pie)

If you haven't drawn a coordinate system showing the point in rectangular coordinates, you should do so. The x-coordinate and y-coordinate of the point determine a right triangle. What quadrant is this triangle in?

Label the base, altitude, and hypotenuse of this triangle with the values of x, y, and r. Then use one of the trig functions to find the angle that the hypotenuse makes with the base. To get the value of theta for your polar point, you will need to take into account which quadrant the triangle is in.

 

[Mark44]

Need answer in the form

(1 , pie/3)

I meant to mention this before. The name of this Greek letter - $\pi$ - is pi, not pie.

 

[imy786]

(3√7)^2 = 63


√ (25 + 63) =

=√88

= √(4 x 22)

=2√22

= 9.38


x co-ordinate = -5
y co-ordinate = 3√7 = 7.94

cos (theta) = -5 / 3√7
theta = 129 degrees

pi = 180 degrees

129 degrees = 0.0243 radians

 

[Mark44]

(3√7)^2 = 63


√ (25 + 63) =

=√88

= √(4 x 22)

=2√22

= 9.38


x co-ordinate = -5
y co-ordinate = 3√7 = 7.94

Up to here, you're OK, although the numbers 9.38 and 7.97 are only rough approximations to 2√22 and 3√7.

cos (theta) = -5 / 3√7

No. The cosine of an angle in a right triangle is the adjacent side divided by the hypotenuse.

theta = 129 degrees



pi = 180 degrees

129 degrees = 0.0243 radians

How did you get this? 129 degrees is a little over 2 radians.