Answer:
u = Cos 60° i + Sin 60° j = (1/2) i + (√3/2) j
v = Cos 30° i - Sin 30° j = (√3/2) i - (1/2) j
Step-by-step explanation:
a) If y = 2sin(x)
y' = m = (2sin (x))' = 2cos(x)
Given
x = π/6
⇒ y' = m = 2cos(π/6) = √3
then we apply
∅ = tan⁻¹(√3) = 60°
If u is the unit vector that lies in the first quadrant that is parallel to the tangent line we have
u = Cos 60° i + Sin 60° j = (1/2) i + (√3/2) j
b) We apply
(1/2, √3/2).(a, -b) = 0
⇒ (a/2) - (√3/2)b = 0
⇒ a = √3b
if b = 1/2
a = √3/2
v = Cos 30° i - Sin 30° j = (√3/2) i - (1/2) j
v is the unit vector that lies in the fourth quadrant that is perpendicular to the tangent line. i + j
a) The unit vector that lies in the first quadrant that is parallel to the tangent line is [tex]\vec u = \frac{1}{4}\,\hat{i} + \frac{\sqrt{3}}{4}\,\hat{j}[/tex].
b) The unit vector that lies in the fourth quadrant that is perpendicular to the tangent line is [tex]\vec u = \frac{\sqrt{3}}{2}\,\hat{i} -\frac{1}{2}\,\hat{j}[/tex].
By calculus we know that that the line tangent to the curve is represented by the first derivative evaluated at the point given on statement:
[tex]m = 2\cdot \cos x[/tex] (1)
[tex]m = 2\cdot \cos \frac{\pi}{6}[/tex]
[tex]m = \sqrt{3}[/tex]
By trigonometry we know that direction of the tangent line is represented by tangent function, then we have the following expression:
[tex]\vec r = \hat{i} + \sqrt{3}\,\hat{j}[/tex] (2)
The unit vector is found by dividing (2) by its magnitude:
[tex]\vec u = \frac{1}{4}\,\hat{i} + \frac{\sqrt{3}}{4}\,\hat{j}[/tex] [tex]\blacksquare[/tex]
The unit vector that lies in the first quadrant that is parallel to the tangent line is [tex]\vec u = \frac{1}{4}\,\hat{i} + \frac{\sqrt{3}}{4}\,\hat{j}[/tex]. [tex]\blacksquare[/tex]
By analytical geometry, the relationship between the line tangent to the curve ([tex]m[/tex]) and the line perpendicular to that ([tex]m_{\perp}[/tex]) is:
[tex]m_{\perp} = -\frac{1}{m}[/tex] (3)
If we know that [tex]m = \sqrt{3}[/tex], then the slope of the line perpendicular to the line:
[tex]m_{\perp} = -\frac{\sqrt{3}}{3}[/tex]
Now we use the same approach taken for the previous point:
[tex]\vec r = \hat{i} - \frac{\sqrt{3}}{3} \,\hat{j}[/tex] (4)
[tex]\vec u = \frac{\sqrt{3}}{2}\,\hat{i} -\frac{1}{2}\,\hat{j}[/tex] [tex]\blacksquare[/tex]
The unit vector that lies in the fourth quadrant that is perpendicular to the tangent line is [tex]\vec u = \frac{\sqrt{3}}{2}\,\hat{i} -\frac{1}{2}\,\hat{j}[/tex]. [tex]\blacksquare[/tex]
To learn more on vectors, we kindly invite to check this verified question: https://brainly.com/question/13188123
