IDNLearn.com is your go-to resource for finding precise and accurate answers. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
Let's convert the vector [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex] to its trigonometric form.
### Step-by-Step Solution:
1. Identify the components [tex]\( t_i \)[/tex] and [tex]\( t_j \)[/tex]:
- The vector given is [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex].
- Here, [tex]\( t_i = -5 \)[/tex] is the [tex]\(i\)[/tex]-component (or the horizontal component).
- [tex]\( t_j = 12 \)[/tex] is the [tex]\(j\)[/tex]-component (or the vertical component).
2. Calculate the magnitude [tex]\( r \)[/tex] of the vector:
- The magnitude [tex]\( r \)[/tex] of a vector [tex]\(\mathbf{t} = t_i\mathbf{i} + t_j\mathbf{j}\)[/tex] is given by:
[tex]\[ r = \sqrt{t_i^2 + t_j^2} \][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[ r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \][/tex]
- Therefore, the magnitude [tex]\( r \)[/tex] of the vector is [tex]\( 13.0 \)[/tex].
3. Calculate the angle [tex]\( \theta \)[/tex] in radians:
- The angle [tex]\( \theta \)[/tex] that the vector makes with the positive [tex]\( x \)[/tex]-axis (the horizontal axis) can be found using the arctangent function:
[tex]\[ \theta = \arctan\left(\frac{t_j}{t_i}\right) \][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{12}{-5}\right) \][/tex]
- The arctangent function will return a value in the correct quadrant where the vector lies. Since [tex]\( t_i \)[/tex] is negative and [tex]\( t_j \)[/tex] is positive, the vector lies in the second quadrant.
- Therefore, the angle [tex]\( \theta \)[/tex] is approximately [tex]\( 1.965587446494658 \)[/tex] radians.
4. Express the vector [tex]\(\mathbf{t}\)[/tex] in trigonometric form:
- The trigonometric form of a vector is given by [tex]\( r (\cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j}) \)[/tex].
- Here, [tex]\( r = 13.0 \)[/tex] and [tex]\( \theta = 1.965587446494658 \)[/tex].
Therefore, the trigonometric form of the vector [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex] is:
[tex]\[ \mathbf{t} = 13.0 (\cos(1.965587446494658) \mathbf{i} + \sin(1.965587446494658) \mathbf{j}) \][/tex]
### Step-by-Step Solution:
1. Identify the components [tex]\( t_i \)[/tex] and [tex]\( t_j \)[/tex]:
- The vector given is [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex].
- Here, [tex]\( t_i = -5 \)[/tex] is the [tex]\(i\)[/tex]-component (or the horizontal component).
- [tex]\( t_j = 12 \)[/tex] is the [tex]\(j\)[/tex]-component (or the vertical component).
2. Calculate the magnitude [tex]\( r \)[/tex] of the vector:
- The magnitude [tex]\( r \)[/tex] of a vector [tex]\(\mathbf{t} = t_i\mathbf{i} + t_j\mathbf{j}\)[/tex] is given by:
[tex]\[ r = \sqrt{t_i^2 + t_j^2} \][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[ r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \][/tex]
- Therefore, the magnitude [tex]\( r \)[/tex] of the vector is [tex]\( 13.0 \)[/tex].
3. Calculate the angle [tex]\( \theta \)[/tex] in radians:
- The angle [tex]\( \theta \)[/tex] that the vector makes with the positive [tex]\( x \)[/tex]-axis (the horizontal axis) can be found using the arctangent function:
[tex]\[ \theta = \arctan\left(\frac{t_j}{t_i}\right) \][/tex]
- Substituting the values [tex]\( t_i = -5 \)[/tex] and [tex]\( t_j = 12 \)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{12}{-5}\right) \][/tex]
- The arctangent function will return a value in the correct quadrant where the vector lies. Since [tex]\( t_i \)[/tex] is negative and [tex]\( t_j \)[/tex] is positive, the vector lies in the second quadrant.
- Therefore, the angle [tex]\( \theta \)[/tex] is approximately [tex]\( 1.965587446494658 \)[/tex] radians.
4. Express the vector [tex]\(\mathbf{t}\)[/tex] in trigonometric form:
- The trigonometric form of a vector is given by [tex]\( r (\cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j}) \)[/tex].
- Here, [tex]\( r = 13.0 \)[/tex] and [tex]\( \theta = 1.965587446494658 \)[/tex].
Therefore, the trigonometric form of the vector [tex]\(\mathbf{t} = -5\mathbf{i} + 12\mathbf{j}\)[/tex] is:
[tex]\[ \mathbf{t} = 13.0 (\cos(1.965587446494658) \mathbf{i} + \sin(1.965587446494658) \mathbf{j}) \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.
