https://file.aiquickdraw.com/mathbot/file/17259908530809n8ghqxh.jpg

Let's work through this step by step:

### Question 8a:
We are given the voltage equation:

\[
V = 55\sqrt{3} \sin\left(\frac{\pi t}{30}\right) + 55 \cos\left(\frac{\pi t}{30}\right)
\]

We need to express this in the form:

\[
R \sin\left(\frac{\pi t}{30} + \alpha \right)
\]

where \( R = 110 \) and \( \alpha = \frac{\pi}{6} \).

---

To transform a function of the form \( A\sin(x) + B\cos(x) \) into \( R\sin(x + \alpha) \), we use the following identities:
1. \( R = \sqrt{A^2 + B^2} \)
2. \( \tan(\alpha) = \frac{B}{A} \)

#### Step 1: Find \( R \)

Here, \( A = 55\sqrt{3} \) and \( B = 55 \). Therefore,

\[
R = \sqrt{(55\sqrt{3})^2 + (55)^2} = \sqrt{55^2 \times 3 + 55^2} = \sqrt{55^2(3 + 1)} = \sqrt{55^2 \times 4} = 55 \times 2 = 110
\]

So, \( R = 110 \), which matches the given value.

#### Step 2: Find \( \alpha \)

We use the relation:

\[
\tan(\alpha) = \frac{B}{A} = \frac{55}{55\sqrt{3}} = \frac{1}{\sqrt{3}}
\]

Since \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), we find that \( \alpha = \frac{\pi}{6} \).

Thus, the expression is indeed of the form \( 110\sin\left(\frac{\pi t}{30} + \frac{\pi}{6}\right) \).

This confirms the required result for 8a.

---

### Question 8b:
Now, for the second part of the question, we need to find the voltage at two specific times:

#### (i) Find the voltage at time \( t = 0 \):

From the expression we just derived:

\[
V(t) = 110 \sin\left(\frac{\pi t}{30} + \frac{\pi}{6}\right)
\]

At \( t = 0 \):

\[
V(0) = 110 \sin\left(\frac{0}{30} + \frac{\pi}{6}\right) = 110 \sin\left(\frac{\pi}{6}\right)
\]

Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \):

\[
V(0) = 110 \times \frac{1}{2} = 55
\]

So, the voltage at \( t = 0 \) is **55 volts**.

#### (ii) Find the voltage after one minute:

One minute corresponds to \( t = 60 \) seconds. So, we substitute \( t = 60 \) into the expression:

\[
V(60) = 110 \sin\left(\frac{\pi \times 60}{30} + \frac{\pi}{6}\right) = 110 \sin\left(2\pi + \frac{\pi}{6}\right)
\]

Since \( \sin(2\pi + \theta) = \sin(\theta) \), this simplifies to:

\[
V(60) = 110 \sin\left(\frac{\pi}{6}\right) = 110 \times \frac{1}{2} = 55
\]

Thus, the voltage after one minute is also **55 volts**.

---

Would you like further details on any of these calculations? Here are 5 additional questions to explore:

1. How would the voltage change if the initial amplitude was different?
2. Can you describe the effect of changing the phase shift \( \alpha \)?
3. How do you interpret the physical meaning of \( R \) in this context?
4. What would happen to the voltage if the time period was halved?
5. How can you express the given voltage function using only cosine?

**Tip:** When converting between sine and cosine, remember that \( \sin(x) = \cos\left(x - \frac{\pi}{2}\right) \).

The alternating voltage, V, in an electrical circuit, t seconds after it is switched on, is modeled by the function V = 55√3 sin(πt/30) + 55 cos(πt/30). Show that this can be written as R sin(πt/30 + α), where R = 110 and α = π/6. Find the voltage at time t = 0 and after one minute.

Understand how to express an alternating voltage function in the form R sin(x + α) using trigonometric identities and formulas. This problem involves voltage modeling, harmonic functions, and phase shifts, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation