Trig Critical Number Problem

[Jason76]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle f'(x) = 9\sin\theta(2 + \cos \theta)\)

\(\displaystyle f'(x) = 9\sin\theta = 0\)

\(\displaystyle f'(x) = (2 + \cos \theta) = 0\)

 

[Deleted member 4993]

f(x) = \(\displaystyle 18 \cos\theta + 9\sin^{2} \theta\)

f'(x) = \(\displaystyle 18\sin\theta + 18\sin \theta \cos \theta\)

f'(x) = \(\displaystyle 9\sin\theta(2 + \cos \theta)\) ← Incorrect .... bad algebra

f'(x) = \(\displaystyle 9\sin\theta = 0\)

.

 

[Jason76]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)


\(\displaystyle 18\sin\theta(1 + \cos \theta)\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle f'(x) = (2 + \cos \theta) = 0\)

 

[Deleted member 4993]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle f'(x) = 18\sin\theta(1 + \cos \theta)\)

\(\displaystyle f'(x) = 18\sin\theta = 0\)

\(\displaystyle f'(x) = (2 + \cos \theta) = 0\) ← How did you get that????

.

 

[Jason76]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle 18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle (1 + \cos \theta) = 0\)

\(\displaystyle \cos \theta = -1\)

\(\displaystyle \theta = \pi\) ??

Critical numbers: \(\displaystyle (0,\pi)\) ??

 

[DrPhil]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)...X derivative of cosine is negative sine

\(\displaystyle 18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle (1 + \cos \theta) = 0\)...correct this because of error above

\(\displaystyle \cos \theta = -1\)

\(\displaystyle \theta = \pi\) ??

Critical numbers: \(\displaystyle (0,pi)\) ??

You need to clarify what you are doing. Setting the derivative to zero implies that one or the other factor is zero, so setting each factor to zero (independently) will reveal all of the critical values.

\(\displaystyle \sin\theta = 0 \implies \theta = \ \cdot \ \cdot \ \cdot \)

other factor \(\displaystyle = 0 \implies \theta = \ \cdot \ \cdot \ \cdot \)

 

[Jason76]

ok, corrected :cool:

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = -18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle -18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(-1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle (-1 + \cos \theta) = 0\)

\(\displaystyle \cos \theta = 1\)

\(\displaystyle \theta = 1\) ??

Critical numbers: \(\displaystyle (0,1)\) ??

 

[Deleted member 4993]

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = 18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle 18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle (1 + \cos \theta) = 0\)

\(\displaystyle \cos \theta = -1\)

\(\displaystyle \theta = \pi\) ??

Critical numbers: \(\displaystyle (0,\pi)\) ??

As DrPhil pointed out before, you should have

sin(Θ) = 0 → Θ = n*\(\displaystyle \pi\) where n = 0, ±1, ±2, ±3....

and

1 - cos(Θ) = 0 → cos(Θ) = 1 → Θ = 2*n*\(\displaystyle \pi\) where n = 0, ±1, ±2, ±3....

 

[Jason76]

As DrPhil pointed out before, you should have

sin(Θ) = 0 → Θ = n*\(\displaystyle \pi\) where n = 0, ±1, ±2, ±3....

and

1 - cos(Θ) = 0 → cos(Θ) = 1 → Θ = \(\displaystyle \dfrac{\pi}{2}\) ± 2*n*\(\displaystyle \pi\) where n = 0, 1, 2, 3....

But the

\(\displaystyle \cos \theta = 1\)

\(\displaystyle \theta = 0\)


not \(\displaystyle \dfrac{\pi}{2}\)

 

[Deleted member 4993]

But the

\(\displaystyle \cos \theta = 1\)

\(\displaystyle \theta = 0\)


not \(\displaystyle \dfrac{\pi}{2}\)

You are correct - I fixed it above

 

[Jason76]

ok, corrected again :cool:

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = -18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle -18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(-1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle \theta = 0\)

\(\displaystyle (-1 + \cos \theta) = 0\)

\(\displaystyle \cos \theta = 1\)

\(\displaystyle \theta = 0\) ??

Critical numbers: \(\displaystyle (0,0)\) ?? - computer still says wrong

 

[Jason76]

Answer is \(\displaystyle n\pi\) You can't type in \(\displaystyle 0\) degrees.

ok, corrected :cool:

\(\displaystyle f(x) = 18 \cos\theta + 9\sin^{2} \theta\)

\(\displaystyle f'(x) = -18\sin\theta + 18\sin \theta \cos \theta\)

\(\displaystyle -18\sin\theta + 18\sin \theta \cos \theta = 0\)


\(\displaystyle 18\sin\theta(-1 + \cos \theta) = 0\)

\(\displaystyle 18\sin\theta = 0\)

\(\displaystyle \sin\theta = 0\)

\(\displaystyle (-1 + \cos \theta) = 0\)

\(\displaystyle \cos \theta = 1\)

\(\displaystyle \theta = 1\) ??

Critical numbers: \(\displaystyle (0,1)\) ??

 

[Deleted member 4993]

Answer is \(\displaystyle n\pi\) You can't type in \(\displaystyle 0\) degrees.

n\(\displaystyle \pi\) = 0 when n = 0

so

0 radian is included in \(\displaystyle n\pi\)