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BAB IV

BAB IV. FUNGSI VEKTOR DALAM RUANG DIMENSI TIGA. 4.1 FUNGSI VEKTOR. Fungsi Vektor dalam ruang dimensi tiga ditentukan oleh r(t) = f(t) i + g(t) j + h(t) k Cara menggambar busur suatu persamaan vektor Substitusi nilai t dalam interval ke persamaan vektor.

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BAB IV

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  1. BAB IV FUNGSI VEKTOR DALAM RUANG DIMENSI TIGA

  2. 4.1 FUNGSI VEKTOR • Fungsi Vektor dalam ruang dimensi tiga ditentukan oleh r(t) = f(t) i + g(t) j + h(t) k • Cara menggambar busur suatu persamaan vektor • Substitusi nilai t dalam interval ke persamaan vektor. • Gambarkan titik-titik tersebut dalam ruang dimensi tiga. • Hubungkan titik-titik tersebut.

  3. 4.2 KECEPATAN, PERCEPATAN, DAN PANJANG BUSUR • Jika fungsi vektor r(t) = f(t) i + g(t) j + h(t) k maka kecepatan = v(t) = r’(t) percepatan = a(t) = r”(t) panjang busur = s

  4. 4.3 KELENGKUNGAN DAN KOMPONEN VEKTOR • Kelengkungan(κ) pada persamaan vektor r(t) = f(t) i + g(t) j + h(t) k di titik t = t1

  5. Komponen Vektor • Vektor Singgung • Vektor Normal

  6. Contoh Soal Kelengkungan Tentukan kelengkungan r = (t2-1)i + (2t+3)j + (t2-4t)k di t = 2 Solusi r’ = 2ti + 2j + (2t-4)k |r’| = 2√2t2-4t+5

  7. T = [ti + j + (t-2)k] [ 2t2 - 4t + 5]-1/2 T = [t (2t2 - 4t + 5)-1/2 i] + [(2t2 - 4t + 5)-1/2 j] + [(t-2) (2t2 - 4t + 5)-1/2 k ] T’ = [(2t2- 4t+5)-1/2 + t(-1/2)(4t-4)(2t2 - 4t + 5)-3/2 ] i + [(-1/2)(4t-4)(2t2 - 4t + 5)-3/2] j + [(2t2- 4t+5)-1/2 + (t-2)(-1/2) (4t-4) (2t2 - 4t + 5)-3/2] k

  8. T’ = [(2t2- 4t+5)-1/2 + (-2t2+2t)(2t2 - 4t + 5)-3/2 ] i + [(-2t+2)(2t2 - 4t + 5)-3/2] j + [(2t2- 4t+5)-1/2+(-2t2+6t-4)(2t2 - 4t + 5)-3/2] k r’ = 2ti + 2j + (2t-4)k r’(t=2) = 4i + 2j |r’| = √20 = 2√5 T’(t=2) = [1/(5√5) i – 2/(5√5) j + 1/√5 k ] |T’| = √1/125 + 4/125 + 1/5 |T’| = √30/125

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