Maxima and Minima Trig Problem

Hello,

I'm having difficulty in understanding where the book has got these angles from:

Question: Find the maximum and minimum values for $\displaystyle y = 3\sin\theta - 4\cos\theta\ in\ the\ range\ \theta\ to\ 2\pi $

$\displaystyle \frac{d}{d\theta} (3\sin\theta - 4\cos\theta) = 4\sin\theta + 3\cos\theta $

$\displaystyle 4\sin\theta + 3\cos\theta = R\sin(\theta + \alpha) $

$\displaystyle 4\sin\theta + 3\cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha $

$\displaystyle R = \sqrt {4^2 + 3^2} = \pm5 $

$\displaystyle \alpha = \arctan \frac{3}{4} \approx 36.869^{\circ} $

$\displaystyle \frac {dy}{d\theta} = 5\sin(\theta + 36.869^{\circ}) = \pm 5\ for\ a\ maximum\ or\ minimum\ value $

$\displaystyle \theta \approx 90 - 36.869^{\circ} \approx 53.131^{\circ}$

Now, in my book the answer is 5 at $\displaystyle 143^{\circ}8' $ for a maximum value, and -5 at $\displaystyle 323^{\circ}8' $ for a minimum value. However, they have subtracted $\displaystyle \alpha $ from $\displaystyle 180^{\circ} $ in order to obtain these pair of angles? I've drawn the function, and these values is where $\displaystyle 5\sin(\theta + 36.869^{\circ}) = 0 $. As a matter of fact, the maximum value first occurs at $\displaystyle 53.131^{\circ} $ and the minimum value at $\displaystyle 233.131^{\circ}$.

I've looked into this a bit deeper, and the only way a maximum or minimum value can occur at these computed angles - is if the first derivative is a negative cosine function.

What have I done wrong?

Thank you for your attention.

Re: Maxima and Minima Trig Problem

first off, the problem states the interval of interest is $\displaystyle 0 \le \theta < 2\pi$ ... you need to start thinking radians.

$\displaystyle y = 3\sin{\theta} - 4\cos{\theta}$

$\displaystyle \frac{dy}{d\theta} = 3\cos{\theta} + 4\sin{\theta} = 0$

for $\displaystyle \frac{dy}{d\theta} = 0$ , $\displaystyle \cos{\theta}$ and $\displaystyle \sin{\theta}$ will have to have opposite signs ... which tells you $\displaystyle \theta$ will have to be in quadrants II or IV.

$\displaystyle \tan{\theta} = -\frac{3}{4}$

$\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + \pi$, a quad II angle

$\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi$, a quad IV angle

to finish, use the second derivative test for the two angles to determine max/min.

Re: Maxima and Minima Trig Problem

Quote:

for $\displaystyle \frac{dy}{d\theta} = 0$ , $\displaystyle \cos{\theta}$ and $\displaystyle \sin{\theta}$ will have to have opposite signs ... which tells you $\displaystyle \theta$ will have to be in quadrants II or IV.

$\displaystyle \tan{\theta} = -\frac{3}{4}$

$\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + \pi$, a quad II angle

$\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi$, a quad IV angle

to finish, use the second derivative test for the two angles to determine max/min.

I don't understand. For $\displaystyle \frac {dy}{d\theta} $ why do we have to give sin and cos opposite signs to equal zero?

If sin and cos are both made positive does this not make $\displaystyle \tan \theta = \frac{-3}{-4} = \frac {3}{4} $ ?

Further, why have you added $\displaystyle \pi $ and $\displaystyle 2\pi $ to $\displaystyle \tan = -\frac{3}{4} $?

Re: Maxima and Minima Trig Problem

Quote:

Originally Posted by

**astartleddeer** I don't understand. For $\displaystyle \frac {dy}{d\theta} $ why do we have to give sin and cos opposite signs to equal zero?

3(something) + 4(something else) = 0

what does that tell you about the signs of (something) and (something else) ?

Quote:

If sin and cos are both made positive does this not make $\displaystyle \tan \theta = \frac{-3}{-4} = \frac {3}{4} $ ?

if sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.

Quote:

Further, why have you added $\displaystyle \pi $ and $\displaystyle 2\pi $ to $\displaystyle \tan = -\frac{3}{4} $?

you need to research the range of the arctangent function.

Re: Maxima and Minima Trig Problem

Quote:

Originally Posted by

**skeeter** 3(something) + 4(something else) = 0

what does that tell you about the signs of (something) and (something else) ?

Then only one of them can be negative?

Quote:

If sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.

Oops. I was meant to say, if sin and cos are both negative then the tangent is positive in the third quadrant?

Quote:

you need to research the range of the arctangent function.

Oh my goodness! I had mistaken $\displaystyle \pi $ for $\displaystyle 90^{\circ} $! Yes, I'm with you now. If the tangent is negative, it is $\displaystyle \pi + (-0.64^{c}) $ or $\displaystyle 2\pi + (-0.64^{c}) $. Second and fourth quadrant respectively.

Re: Maxima and Minima Trig Problem

Ok let's do this.

$\displaystyle \frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value $

$\displaystyle R = \sqrt{4^2 + (-3)^2} = \pm 5 $

$\displaystyle \theta = \arctan -\frac{3}{4} \approx -0.6435^{c} $

$\displaystyle In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{c} $ and $\displaystyle 2\pi + (-0.6435^{c}) \approx 5.639^{c} $

$\displaystyle \frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta $

$\displaystyle \frac{d^{2}y}{d\theta^2} $ is a maximum for $\displaystyle 2.496^{c} $

$\displaystyle \frac{d^{2}y}{d\theta^2} $ is a minimum for $\displaystyle 5.639^{c} $

Therefore, a minimum value occurs at $\displaystyle -5, 5.639^{c} $ and a maximum value occurs at $\displaystyle 5, 2.496^{c} $

Re: Maxima and Minima Trig Problem

Quote:

Originally Posted by

**astartleddeer** Ok let's do this.

[tex] \frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value

$\displaystyle R = \sqrt{4^2 + (-3)^2} = \pm 5 $

$\displaystyle \theta = \arctan = -\frac{3}{4} \approx -0.6435^{\circ} $

$\displaystyle In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{\circ} $ and $\displaystyle 2\pi + (-0.6435^{c}) \approx 5.639^{c} $

$\displaystyle \frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta $

$\displaystyle \frac{d^{2}y}{d\theta^2} $ is a maximum for $\displaystyle 2.496^{c} $

$\displaystyle \frac{d^{2}y}{d\theta^2} $ is a minimum for $\displaystyle 5.639^{c} $

Therefore, a minimum value occurs at $\displaystyle -5, 5.639^{c} $ and a maximum value occurs at $\displaystyle 5, 2.496^{c} $

Please confirm this.

maximum function value of $\displaystyle y = 5$ at $\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + \pi \approx 2.498$ **RADIANS**

minimum function value of $\displaystyle y = -5$ at $\displaystyle \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi \approx 5.640$ **RADIANS**

Re: Maxima and Minima Trig Problem

I indicated radians with a small c for circular measure (Happy)

Ok, thank you for your time Skeeter. Done and dusted (Clapping)