Question

(+50) KuAT – Kuis Awal Tahun

Garis [tex]g[/tex] dan [tex]h[/tex] adalah garis singgung kurva fungsi [tex]y = \sin(x) + \cos(x)[/tex] pada interval 0 ≤ [tex]x[/tex] ≤ 2π. Masing-masing garis tersebut tegak lurus terhadap garis [tex]y = x[/tex].
Berapakah jarak antara garis [tex]g[/tex] dan [tex]h[/tex]?

Answer 1

Verified answer

Jawab:
= ¼ · (4 - π)√2 satuan

Penjelasan:
g dan h garis lurus yg menyinggung
y = sin(x) + cos(x)  {0 ≤ x ≤ 2π}

g dan h tegak lurus terhadap y = x
y = 1x (sesuai dengan y = mx + k, maka)
m = 1
m × mh = -1
1 × mh = -1
-->  mh = -1

y = sin(x) + cos(x)
-->  y' = cos(x) - sin(x)

mh = cos(x) - sin(x)
-1 = cos(x) - sin(x)
cos(x) = 1 - sin(x)
cos²(x) = (1 - sin(x))²
cos²(x) = 1 - 2sin(x) + sin²(x)
cos²(x) = 1 - 2sin(x) + sin²(x)
∵ sin²(x) + cos²(x) = 1 ∴
1 - sin²(x) = 1 - 2sin(x) + sin²(x)
1 - sin²(x) - sin²(x) = 1 - 2sin(x)
1 - 2sin²(x) = 1 - 2sin(x)
2sin²(x) = 2sin(x)
2sin²(x) - 2sin(x) = 0
Subs u dengan sin(x) maka
2u² - 2u = 0
2u(u-1) = 0
u = 0, u = 1
sin(x) = 0, sin(x) = 1
x = 0, π    x = π/2

Cek nyinggung atau enggak
-1 = cos(x) - sin(x)
-1 = cos(0) - sin(0)
-1 = 1 - 0 (tidak memenuhi)

-1 = cos(x) - sin(x)
-1 = cos(π) - sin(π)
-1 = -1 - 0 (memenuhi)

-1 = cos(x) - sin(x)
-1 = cos(π/2) - sin(π/2)
-1 = 0 - 1 (memenuhi)

Artinya
x = {π, π/2}

Garis g
Untuk x₁ = π
y₁ = sin(x₁) + cos(x₁)
y₁ = sin(π) + cos(π)
y₁ = 0 + (-1)
y₁ = -1
Persamaan garis g
y - y₁ = (mh)(x - x₁)
y - (-1) = -1(x - π)
y + 1 = -x + π
-->  x + y + 1 - π = 0

Garis h
Untuk x₁ = π/2
y₁ = sin(x₁) + cos(x₁)
y₁ = sin(π/2) + cos(π/2)
y₁ = 1 + 0
y₁ = 1
Persamaan garis h
y - y₁ = (mh)(x - x₁)
y - 1 = -1(x - (π/2))
y - 1 = -x + (π/2)
y = -x + (π/2) + 1
-->  x + y - 1 - (π/2) = 0

Gabungkan
g   -->   x + y + 1 - π = 0
A = 1, B = 1, C = 1 - π
h   -->   x + y - 1 - (π/2) = 0
..              D = - 1 - (π/2)

Jarak garis g dan h
= (| C - D |) / √(A²+B²)
= (|1 - π - (- 1 - (π/2))|) / √(1²+1²)
= (|1 - π + 1 + (π/2)|) / √2
= (|2 - π + (π/2)|) / √2
= (|(4/2) - (2π/2) + (π/2))|) / √2
= (|4 - 2π + π|) / (2√2)
= (|4 - π|) / (2√2)
= (4 - π)√2 / (2√2√2)
= (4 - π)√2 / (2(2))
= (4 - π)√2 / 4 satuan
= ¼ · (4 - π)√2 satuan

(xcvi)

Answer 2

Aplikasi Turunan

garis 1 : y = x tegak lurus garis 2 : y = -x + c

y = sin x + cos x

y = k cos (x - a)

y = √2 cos (x - π/4)

y' = m2 = -1

- √2 sin (x - π/4) = -1

sin (x - π/4) = 1/2 √2

x - π/4 = π/4 + k.2π

x = π/2 + k.2π ... (1)

atau

x - π/4 = 3π/4 + k.2π

x = π + k.2π ... (2)

titik singgung :

x1 = π/2 → y1 = sin 90° + cos 90° = 1

x2 = π → y2 = -1

(π , -1)

(π/2 , 1)__(-)

(π/2 , -2)

PGS :

y = -x + c

x + y = c

x + y = mx1 + y1

a = 1 ; b = 1

Jarak kedua garis singgung

= |c1 - c2| / √(a² + b²)

= |m(x1 - x2) - (y1 - y2)| / √2

= |-π/2 - (-2)| / √2

= 1/4 (4 - π)√2 satuan