Chapter 13.5, Problem 63HP

To determine

To find: how many solutions in the interval $0^{\circ }\le \theta \le {360}^{\circ }$ should expect for

  $a\sin (b\theta +c)+d=d\left(\frac{a}{2}\right),ifa\ne 0\text{and}b\text{is}\text{a}\text{positive}\text{integer}$ .

Answer to Problem 63HP

If $\left|d\right|>\left|a\right|$ then there are no solutions.

If $\left|d\right|=\left|a\right|$ , b solutions or b + 1 if there is a solution at $\theta =0$ .

If $\left|d\right|<\left|a\right|$ , 2b solutions or 2b + 1 if there is a solution at $\theta =0$ .

Explanation of Solution

Given information: Given equation is; $a\sin (b\theta +c)+d=d\left(\frac{a}{2}\right),0^{\circ }\le \theta \le {360}^{\circ }$

Calculation:

  $a\sin (b\theta +c)+d=d\left(\frac{a}{2}\right),0^{\circ }\le \theta \le {360}^{\circ }$ .

It helps a lot that b is restricted to a positive integer because otherwise it would be much more complicated. The period is found by first factoring:

  $\begin{array}{l}b\theta +c=b(\theta +\frac{c}{b})\\ period=\frac{{360}^{\circ }}{b}\end{array}$

So since b is a positive integer, then there is one complete cycle when b = 1, 2 complete cycle when b =2, 3 complete cycle when b = 3 etc.

The place shift is $\frac{c}{b}$ , which can almost mostly ignore because any amount of the graph shift off one side of the interval will reappear on the other side. It becomes significant for the case when a solution to y = d occurs when $\theta =0$ . Then there is an additional solution where $\theta ={360}^{\circ }$.

The only thing a does is vertical stretch. So then;

If $\left|d\right|>\left|a\right|$ then there are no solutions for y =d , d is outside the range of the function.

If $\left|d\right|=\left|a\right|$ then count how many maximum points there are. 1 cycle has 1 max point. 2 cycles has 2 points, etc., so then in this case the number of solutions is b . If a solution is at $\theta =0$ then the number of solutions is b +1.

If $\left|d\right|<\left|a\right|$ then count how many times the graph increases to and from a maximum point. 1 cycle has 2 of these, 2 cycles has 4. etc. so then in this case the number of solution is

2b . If a solution is at $\theta =0$ then the number of solutions is 2b +1.

The graph is shown in radians (0, to $2\pi$ ). With a = 1 and b = 3 so there are 3 complete cycles and phase shift so that a solution

For d =0.5 is at $\theta =0.$ In that case there are 2b +1=2(3) +1=7 solutions.

For d =-0.5 is at $\theta =0.$ In that case there are 2b =2(3) =6 solutions.

For d =1 is at $\theta =0.$ In that case there are 2b +1=2(1) +1=3 solutions.

  

Therefore,

If $\left|d\right|>\left|a\right|$ then there are no solutions.

If $\left|d\right|=\left|a\right|$ , b solutions or b + 1 if there is a solution at $\theta =0$ .

If $\left|d\right|<\left|a\right|$ , 2b solutions or 2b + 1 if there is a solution at $\theta =0$ .