Test for symmetry with respect to Q=pi/2, the polar axis, and the pole.
r=6 sinQ
I know I have to replace (r,Q) with (r,pi-Q) (r,-Q) and (-r,Q) so would that be (6,pi/2)? I do not know how to find the Q.
To test for symmetry with respect to Q=pi/2 (the vertical axis), you will substitute (r,Q) with (-r,Q) and check if the equation remains the same.
Given the equation r=6sinQ, we will substitute (-r,Q) into the equation:
(-r) = 6sinQ
Since we are considering symmetry with respect to Q=pi/2, substitute Q=pi/2 into the equation:
(-r) = 6sin(pi/2)
Simplifying the equation:
(-r) = 6(1)
-r = 6
Therefore, we can see that substituting (-r,Q) gives us the equation -r=6, which is not the same as the original equation r=6sinQ. Thus, the graph of r=6sinQ does not exhibit symmetry with respect to Q=pi/2.
To determine whether there is symmetry with respect to the polar axis (Q=pi), you'll substitute (r,Q) with (r,-Q) and check if the equation remains the same.
Substituting (r,-Q) into the equation r=6sinQ, we get:
r = 6sin(-Q)
Since we are testing symmetry with respect to the polar axis, substitute Q=pi into the equation:
r = 6sin(-pi)
Simplifying the equation:
r = 6(0)
r = 0
Therefore, substituting (r,-Q) gives us the equation r=0, which is the same as the original equation r=6sinQ. Thus, the graph of r=6sinQ exhibits symmetry with respect to the polar axis.
To determine whether there is symmetry with respect to the pole, you'll substitute (r,Q) with (-r,-Q) and check if the equation remains the same.
Substituting (-r,-Q) into the equation r=6sinQ, we get:
(-r) = 6sin(-Q)
Since we are testing symmetry with respect to the pole, substitute Q=0 into the equation:
(-r) = 6sin(0)
Simplifying the equation:
(-r) = 6(0)
(-r) = 0
r = 0
Therefore, substituting (-r,-Q) gives us the equation r=0, which is the same as the original equation r=6sinQ. Thus, the graph of r=6sinQ exhibits symmetry with respect to the pole.
To summarize:
- The graph of r=6sinQ does not exhibit symmetry with respect to Q=pi/2 (the vertical axis).
- The graph of r=6sinQ exhibits symmetry with respect to the polar axis (Q=pi).
- The graph of r=6sinQ exhibits symmetry with respect to the pole.
To test for symmetry with respect to Q = pi/2, the polar axis, and the pole, you need to replace (r, Q) with (r, pi - Q), (r, -Q), and (-r, Q) as you mentioned.
Let's start with replacing (r, Q) with (r, pi - Q). In this case, the equation becomes:
r = 6sin(pi - Q)
Now, we know that sin(pi - Q) is equal to sin(pi)cos(Q) - cos(pi)sin(Q) which simplifies to -sin(Q). Therefore, the equation becomes:
r = 6(-sin(Q))
This implies that the graph of the equation will be symmetrical with respect to the polar axis (Q = pi/2) if the negative sign (-) is present.
Next, let's replace (r, Q) with (r, -Q). The equation becomes:
r = 6sin(-Q)
Using the identity sin(-Q) = -sin(Q), the equation simplifies to:
r = -6sin(Q)
Again, we have a negative sign, indicating that the graph will be symmetrical with respect to the polar axis.
Finally, let's replace (r, Q) with (-r, Q). The equation becomes:
-r = 6sin(Q)
If we multiply both sides of the equation by (-1), we get:
r = -6sin(Q)
Again, we have a negative sign, indicating symmetry with respect to the pole.
In conclusion, the equation r = 6sin(Q) exhibits symmetry with respect to Q = pi/2 (the polar axis) and the pole. To find specific values of Q for which symmetry occurs, you would need to substitute different Q values into the equation and observe if the resulting graphs are symmetrical.