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Question: How do you find the period and amplitude of \(y = - 3\sin \left( {\dfrac{x}{3}} \right)\) ?...

How do you find the period and amplitude of y=−3sin⁡(x3)y = - 3\sin \left( {\dfrac{x}{3}} \right) ?

Explanation

Solution

Normalize the function and find out the coefficient and substitute in formula. From the given equation, we need to compare it to the normalized sine function equation of y = A sin(B(x + C)) + Dy{\text{ }} = {\text{ }}A{\text{ }}sin\left( {B\left( {x{\text{ }} + {\text{ }}C} \right)} \right){\text{ }} + {\text{ }}D. If it's in the required form, no need for simplification. Then later, we identify the value of B which is a coefficient of x, which will be used in the period formula for the given sine function, for which the formula is Period=2π∣B∣Period = \dfrac{{2\pi }}{{\left| B \right|}} and Amplitude=∣A∣Amplitude = \left| A \right|.

Complete step-by-step solution:
The given equation we have is,
⇒y=−3sin⁡(x3)\Rightarrow y = - 3\sin \left( {\dfrac{x}{3}} \right)
The given equation is of the form y=A⋅sin(Bx)y = A \cdot sin(Bx)
Since the formula for period is Period=2π∣B∣Period = \dfrac{{2\pi }}{{\left| B \right|}}and Amplitude=∣A∣Amplitude = \left| A \right|, we need to find A and B, which are the required coefficients.
So, we compare the given equation with the normal equation to find the value of B, from which the value of A is -3 and B is 13\dfrac{1}{3}.
The next step is to replace “B” with its value in the formula, which gives us
⇒Period=2π∣13∣\Rightarrow Period = \dfrac{{2\pi }}{{\left| {\dfrac{1}{3}} \right|}}
⇒Period=2π(3)\Rightarrow Period = 2\pi (3)
⇒Period=6π\Rightarrow Period = 6\pi
The 3 in the denominator gets shifted to the numerator and finally we get the answer as below.
⇒Period=2π(3)\Rightarrow Period = 2\pi (3)
⇒Period=6π\Rightarrow Period = 6\pi
Now, for the amplitude, we should replace the “A” in the formula, which gives us
Amplitude=∣−3∣Amplitude = \left| { - 3} \right|
Finally, finally we get the amplitude as 33.

Thus the answer are period is 6π6\pi and amplitude is ∣−3∣\left| { - 3} \right|

Note: It is necessary to have the equation be normalized first, as it makes it easy for identifying the coefficients of A, B, C and so on. And also, we should be careful on not missing out the coefficients which are in fraction forms as well as the nature of sign of the coefficient. We can forget to consider denominators as a part of the coefficient value. We should always remember that period and amplitude is always positive.