Ngā Tauira o ngā Pātai e Matapaki ana i te Whakamahinga o ngā Pānga Taurite i roto i ngā Mara Pūtaiao Rerekē
Pendahuluan
He ariā taketake te taupū i roto i te tātaitai me ngā tono maha i roto i ngā momo mara pūtaiao me te hangarau. E whakaahua ana i te tere o te huringa o tētahi mahi, ā, ka taea te whakamahi hei tautuhi i te mōrahi me te iti rawa, hei whakaoti rapanga arotau, hei tātari i te tipu me te hekenga i roto i tētahi pūnaha. I roto i tēnei tuhinga, ka matapakihia e mātou ētahi tauira rapanga me ngā tono o ngā taupū i roto i ngā momo mara pūtaiao, pērā i te ahupūngao, te ōhanga, te koiora, me te hangarau.
1. Ahupūngao: Te Whakaterenga hei Pūtake o te Tere
I roto i te ahupūngao, ko te whakaterenga te pānga o te tere e pā ana ki te wā. Ko tētahi tauira raruraru i roto i tēnei horopaki ko:
Tauira raruraru:
Ka neke tetahi mea i te rārangi tika me te mahi tūnga \(s(t) = 4t^3 – 3t^2 + 2t – 1\) mita, ko \(t\) kei roto i ngā hēkona. Tātaihia te tere me te whakaterenga o te mea i \(t = 2\) hēkona.
Kōrero:
Ko te tere te pānga tuatahi o te tūranga e pā ana ki te wā:
\[ v(t) = s'(t) = \frac{d}{dt}(4t^3 – 3t^2 + 2t – 1) \]
\[ v(t) = 12t^2 – 6t + 2 \]
Ko te whakaterenga te pānga tuatahi o te tere e pā ana ki te wā:
\[ a(t) = v'(t) = \frac{d}{dt}(12t^2 – 6t + 2) \]
\[ a(t) = 24t – 6 \]
Nō reira, ko te tere i te \(t = 2\) hēkona ko:
\[ v(2) = 12(2)^2 – 6(2) + 2 = 48 – 12 + 2 = 38 \, \kuputuhi{m/s} \]
Ko te whakaterenga i te \(t = 2\) hēkona ko:
\[ a(2) = 24(2) – 6 = 48 – 6 = 42 \, \kuputuhi{m/s}^2 \]
2. Ōhanga: Te Arotau i te Hua
I roto i te ōhanga, he maha ngā wā ka whakamahia ngā pānga tauhohenga hei whakatau i ngā pūwāhi mōrahi, mōrahi rānei o tētahi mahi hua, mahi utu rānei. Ko tētahi tauira raruraru i roto i tēnei horopaki ko:
Tauira raruraru:
He kamupene e whakaputa ana i ngā taonga me te mahi hua \(P(x) = -2x^2 + 12x – 20\), ko \(x\) te maha o ngā waeine i hangaia me te hoko. E hia ngā waeine me whakaputa hei whakanui ake i ngā hua, ā, he aha te hua mōrahi?
Kōrero:
Hei whakanui ake i ngā hua, me kimi e tātou te taupū tuatahi o \(P(x)\) me ōna pūwāhi matua.
\[ P'(x) = \frac{d}{dt}(-2x^2 + 12x – 20) \]
\[ P'(x) = -4x + 12 \]
Kimihia te pūwāhi matua mā te whakaoti rapanga \(P'(x) = 0\):
\[ -4x + 12 = 0 \]
\[ x = 3 \]
Me whakarite kia noho a \(x = 3\) hei pūwāhi mōrahi mā te whakamahi i te tuarua o ngā taupū.
\[ P”(x) = \frac{d}{dt}(-4x + 12) \]
\[ P”(x) = -4 \]
Mai i te mea ko te \(P”(3) = -4 < 0\), e whakaatu ana tēnei ko te \(x = 3\) te pūwāhi mōrahi. Ko te hua mōrahi ko: \[ P(3) = -2(3)^2 + 12(3) - 20 \] \[ P(3) = -18 + 36 - 20 \] \[ P(3) = -2 \] Nō reira, ko te maha o ngā waeine me whakaputa hei whakanui i te hua ko te 3 waeine, ā, ko te hua mōrahi ko te -2. 3. Koiora: Te Tere Tipu Taupori I roto i te koiora, ka whakamahia ngā tātaitanga hei tātari i ngā tere tipu taupori. He tauira raruraru i roto i tēnei horopaki ko: Tauira Raru: Me kī ko te rahi o te taupori o tētahi momo ka hoatuhia e te mahi \(P(t) = 100e^{0.05t}\), ko te \(t\) te wā i roto i ngā tau. Kimihia te tere tipu taupori i \(t = 10\). Otinga: Ko te tere tipu taupori ko te tātaitanga o \(P(t)\) e pā ana ki te wā: \[ P'(t) = \frac{d}{dt}(100e^{0.05t}) \] \[ P'(t) = 100 \cdot 0.05e^{0.05t} \] \[ P'(t) = 5e^{0.05t} \] Ko te tere tipu o te taupori i \(t = 10\) ko: \[ P'(10) = 5e^{0.05(10)} \] \[ P'(10) = 5e^{0.5} \] Mā te tatau i te uara taupū o \(e^{0.5}\) (tata ki te 1.64872): \[ P'(10) \approx 5 \cdot 1.64872 \] \[ P'(10) \approx 8.2436 \] Nō reira, ko te tere tipu o te taupori i \(t = 10\) tata ki te 8.24 tāngata ia tau. 4. Hangarau: Te Hoahoa Tino Pai o ngā Ara Hiko I roto i te hangarau, inā koa te hangarau hiko, ka whakamahia ngā momo whakawhiti hei arotau i te hoahoa ara iahiko. Ko tētahi tauira raruraru i tēnei horopaki ko: Tauira Raru: Ki te hoatu he mahi whakapaunga hiko \(P(R) = V^2 / R + I^2 R\), ko \(V\ he ngaohiko pumau, ko \(I\) he iahiko pumau, ā, ko \(R\) he ātete. Whakatauhia te uara o \(R\) e whakaiti ana i te whakapaunga hiko. Otinga: Ko te momo whakawhiti tuatahi o \(P(R)\) e pā ana ki \(R\) ko: \[ P'(R) = \frac{d}{dR}\left(\frac{V^2}{R} + I^2 R\right) \] \[ P'(R) = -\frac{V^2}{R^2} + I^2 \] Hei kimi i te uara o \(R\) e whakaiti ana i te whakapaunga hiko, ka kitea e mātou \(P'(R) = 0\): \[ -\frac{V^2}{R^2} + I^2 = 0 \] \[ \frac{V^2}{R^2} = I^2 \] \[ R^2 = \frac{V^2}{I^2} \] \[ R = \frac{V}{I} \] Nō reira, ko te uara ātete \(R\) e whakaiti ana i te whakapaunga hiko ko \(R = \frac{V}{I}\). Whakamutunga Mai i ngā tauira i runga ake nei, kua kite tātou i te whakamahinga o te ariā o ngā pānga i roto i ngā momo mara pūtaiao pēnei i te ahupūngao, te ōhanga, te koiora, me te miihini. Mā te mārama hohonu ki ngā pānga me ā rātou tono ka taea e tātou te whakaoti rapanga uaua me te arotau i ngā pūnaha i te ao tūturu. He taputapu tātari tino kaha ngā pānga, ā, he whai hua ki te mārama ki ngā hihiri me ngā huringa i roto i ngā horopaki rerekē.
