13.2: Calculus ya Kazi za Vector-Thamani

Malengo ya kujifunza

• Andika maneno kwa derivative ya kazi yenye thamani ya vector.
• Kupata vector tangent katika hatua kwa ajili ya kupewa nafasi vector.
• Kupata kitengo tangent vector katika hatua kwa ajili ya kupewa nafasi vector na kueleza umuhimu wake.
• Tumia sehemu muhimu ya kazi yenye thamani ya vector.

Ili kujifunza calculus ya kazi za thamani ya vector, tunafuata njia sawa na ile tuliyochukua katika kusoma kazi halisi ya thamani. Kwanza, tunafafanua derivative, kisha tunachunguza maombi ya derivative, kisha tunaendelea kufafanua integrals. Hata hivyo, tutapata mawazo mapya ya kuvutia njiani kama matokeo ya asili ya vector ya kazi hizi na mali ya curves nafasi.

Derivatives ya Vector-Thamani Kazi

Sasa kwa kuwa tumeona kazi ya thamani ya vector na jinsi ya kuchukua kikomo chake, hatua inayofuata ni kujifunza jinsi ya kutofautisha kazi yenye thamani ya vector. Ufafanuzi wa derivative ya kazi yenye thamani ya vector ni karibu sawa na ufafanuzi wa kazi halisi ya thamani ya kutofautiana moja. Hata hivyo, kwa sababu aina mbalimbali ya kazi yenye thamani ya vector ina vectors, sawa ni kweli kwa aina mbalimbali za derivative ya kazi yenye thamani ya vector.

Ufafanuzi: Derivative ya Vector-Thamani Kazi

Derivative ya kazi ya thamani ya vector\(\vecs{r}(t)\) ni

\[\vecs{r}′(t) = \lim \limits_{\Delta t \to 0} \dfrac{\vecs{r}(t+\Delta t)−\vecs{r}(t)}{ \Delta t} \label{eq1} \]

zinazotolewa kikomo ipo. Ikiwa\(\vecs{r}'(t)\) ipo, basi\(\vecs{r}(t)\) ni tofauti katika\(t\). Ikiwa\(\vecs{r}′(t)\) ipo kwa wote\(t\) katika kipindi cha wazi\((a,b)\) basi\(\vecs{r}(t)\) ni tofauti zaidi ya muda\((a,b)\). Kwa kazi ili kutofautishwa juu ya muda uliofungwa\([a,b]\), mipaka miwili ifuatayo lazima iwepo pia:

\[\vecs{r}′(a) = \lim \limits_{\Delta t \to 0^+} \dfrac{\vecs{r}(a+\Delta t)−\vecs{r}(a)}{ \Delta t} \nonumber \]

na

\[\vecs{r}′(b) = \lim \limits_{\Delta t \to 0^-} \dfrac{\vecs{r}(b+\Delta t)−\vecs{r}(b)}{ \Delta t} \nonumber \]

Sheria nyingi za kuhesabu derivatives ya kazi halisi za thamani zinaweza kutumika kwa kuhesabu derivatives ya kazi za thamani ya vector pia. Kumbuka kwamba derivative ya kazi halisi ya thamani inaweza kutafsiriwa kama mteremko wa mstari wa tangent au kiwango cha instantaneous cha mabadiliko ya kazi. Derivative ya kazi yenye thamani ya vector inaweza kueleweka kuwa kiwango cha instantaneous cha mabadiliko pia; kwa mfano, wakati kazi inawakilisha nafasi ya kitu kwa wakati fulani, derivative inawakilisha kasi yake kwa hatua hiyo hiyo kwa wakati.

Sasa tunaonyesha kuchukua derivative ya kazi yenye thamani ya vector.

Mfano\(\PageIndex{1}\): Finding the Derivative of a Vector-Valued Function

Tumia ufafanuzi wa kuhesabu derivative ya kazi

\[\vecs{r}(t)=(3t+4) \,\mathbf{\hat{i}}+(t^2−4t+3) \,\mathbf{\hat{j}} .\nonumber \]

Suluhisho

Hebu kutumia Equation\ ref {eq1}:

\ [kuanza {align*}\ vecs {r} (t) &=\ lim\ mipaka _ {\ Delta t\ kwa 0}\ dfrac {\ vecs {r} (T+Δt) -\ vecs {r} (t)} {Δt}\\ [4pt]
&=\ lim\ mipaka _ {\ Delta t\ kwa 0}\ dfrac [{(3 (T+Δt) +4)\,\ kofia {\ mathbf {i}} + (T+Δt) ^2,14 (T+Δt) +3)\,\ kofia {\ hisabati {j}}] - [(3t+4)\,\ kofia {\ mathbf {i}} + (t ^ 2,14t+3)\,\ kofia {\ mathbf {j}}}} {Δt}\\ [4 pt]
&=\ lim\ mipaka _ {\ Delta t\ kwa 0}\ dfrac {(3T+3Δt+4)\,\ kofia {\ mathbf {i}}} - (3t+4)\,\ kofia {\ mathbf {i}} + (T ^ 2+2TΔt+ (Δt) ^2,14t-4Δt+3)\,\ kofia {\ mathbf {j}}} - (t ^ 2,14t+3)\,\ kofia {\ mathbf {j}}} {Δt}\\ [4pt]
&=\ lim\ mipaka _ {\ Delta t\ hadi 0}\ dfrac {(3Δt)\,\ kofia {\ mathbf {i}} + (2TΔtt)\\ kofia {\ mathbf {i}} + (2TΔt)\ + (Δt) ^2—4Δt)\,\ kofia {\ mathbf {j}}} {Δt}\\ [4pt]
&=\ lim\ mipaka _ {\ Delta t\ kwa 0} (3\,\ kofia {\ mathbf {i}} + (2T+Δt-4)\,\ kofia {\ mathbf {j})\\ [4pt]
&=3\,\ kofia {\ mathbf {j} f {i}} + (2t-4)\,\ kofia {\ mathbf {j}}\ mwisho {align*}\ nonumber\]

Zoezi\(\PageIndex{1}\)

Tumia ufafanuzi wa kuhesabu derivative ya kazi\(\vecs{r}(t)=(2t^2+3) \,\mathbf{\hat{i}}+(5t−6) \,\mathbf{\hat{j}}\).

Kidokezo

Tumia Equation\ ref {eq1}.

Jibu

\[\vecs{r}′(t)=4t \,\mathbf{\hat{i}}+5 \,\mathbf{\hat{j}} \nonumber \]

Kumbuka kwamba katika mahesabu katika Mfano\(\PageIndex{1}\), tunaweza pia kupata jibu kwa kuhesabu kwanza derivative ya kila kazi ya sehemu, kisha kuweka derivatives hizi nyuma katika kazi vector-thamani. Hii daima ni kweli kwa kuhesabu derivative ya kazi yenye thamani ya vector, ikiwa ni katika vipimo viwili au vitatu. Tunasema hili katika theorem ifuatayo. Ushahidi wa theorem hii ifuatavyo moja kwa moja kutoka kwa ufafanuzi wa kikomo cha kazi yenye thamani ya vector na derivative ya kazi yenye thamani ya vector.

Theorem\(\PageIndex{1}\): Differentiation of Vector-Valued Functions

Hebu\(f\),\(g\), na\(h\) kuwa differentiable kazi ya\(t\).

1. Kama\(\vecs{r}(t)=f(t) \,\mathbf{\hat{i}}+g(t) \,\mathbf{\hat{j}}\) basi\[\vecs{r}′(t)=f′(t) \,\mathbf{\hat{i}}+g′(t) \,\mathbf{\hat{j}}. \nonumber \]
2. Kama\(\vecs{r}(t)=f(t) \,\mathbf{\hat{i}}+g(t) \,\mathbf{\hat{j}} + h(t) \,\mathbf{\hat{k}}\) basi\[\vecs{r}′(t)=f′(t) \,\mathbf{\hat{i}}+g′(t) \,\mathbf{\hat{j}} + h′(t) \,\mathbf{\hat{k}}. \nonumber \]

Mfano\(\PageIndex{2}\): Calculating the Derivative of Vector-Valued Functions

Tumia Theorem\(\PageIndex{1}\) kuhesabu derivative ya kila moja ya kazi zifuatazo.

1. \(\vecs{r}(t)=(6t+8) \,\mathbf{\hat{i}}+(4t^2+2t−3) \,\mathbf{\hat{j}}\)
2. \(\vecs{r}(t)=3 \cos t \,\mathbf{\hat{i}}+4 \sin t \,\mathbf{\hat{j}}\)
3. \(\vecs{r}(t)=e^t \sin t \,\mathbf{\hat{i}}+e^t \cos t \,\mathbf{\hat{j}}−e^{2t} \,\mathbf{\hat{k}}\)

Suluhisho

Tunatumia Theorem\(\PageIndex{1}\) na kile tunachojua kuhusu kutofautisha kazi za kutofautiana moja.

1. Sehemu ya kwanza ya\[\vecs r(t)=(6t+8) \,\mathbf{\hat{i}}+(4t^2+2t−3) \,\mathbf{\hat{j}} \nonumber \] ni\(f(t)=6t+8\). Sehemu ya pili ni\(g(t)=4t^2+2t−3\). Tuna\(f′(t)=6\) na\(g′(t)=8t+2\), hivyo Theorem\(\PageIndex{1}\) inatoa\(\vecs r′(t)=6 \,\mathbf{\hat{i}}+(8t+2)\,\mathbf{\hat{j}}\).
2. Sehemu ya kwanza ni\(f(t)=3 \cos t\) na sehemu ya pili ni\(g(t)=4 \sin t\). Tuna\(f′(t)=−3 \sin t\) na\(g′(t)=4 \cos t\), hivyo sisi kupata\(\vecs r′(t)=−3 \sin t \,\mathbf{\hat{i}}+4 \cos t \,\mathbf{\hat{j}}\).
3. Sehemu ya kwanza ya\(\vecs r(t)=e^t \sin t \,\mathbf{\hat{i}}+e^t \cos t \,\mathbf{\hat{j}}−e^{2t} \,\mathbf{\hat{k}}\) ni\(f(t)=e^t \sin t\), sehemu ya pili ni\(g(t)=e^t \cos t\), na sehemu ya tatu ni\(h(t)=−e^{2t}\). Tuna\(f′(t)=e^t(\sin t+\cos t)\),\(g′(t)=e^t (\cos t−\sin t)\), na\(h′(t)=−2e^{2t}\), hivyo theorem inatoa\(\vecs r′(t)=e^t(\sin t+\cos t)\,\mathbf{\hat{i}}+e^t(\cos t−\sin t)\,\mathbf{\hat{j}}−2e^{2t} \,\mathbf{\hat{k}}\).

Zoezi\(\PageIndex{2}\)

Tumia derivative ya kazi

\[\vecs{r}(t)=(t \ln t)\,\mathbf{\hat{i}}+(5e^t) \,\mathbf{\hat{j}}+(\cos t−\sin t) \,\mathbf{\hat{k}}. \nonumber \]

Kidokezo

Tambua kazi za sehemu na utumie Theorem\(\PageIndex{1}\).

Jibu

\[\vecs{r}′(t)=(1+ \ln t) \,\mathbf{\hat{i}}+5e^t \,\mathbf{\hat{j}}−(\sin t+\cos t)\,\mathbf{\hat{k}} \nonumber \]

Tunaweza kupanua kwa kazi za thamani ya vector mali ya derivative ambayo tuliwasilisha hapo awali. Hasa, utawala wa mara kwa mara, sheria za jumla na tofauti, utawala wa bidhaa, na utawala wa mnyororo wote hupanua kwa kazi za thamani ya vector. Hata hivyo, katika kesi ya utawala wa bidhaa, kuna kweli upanuzi tatu:

1. kwa kazi halisi yenye thamani tele kwa kazi vector-thamani,
2. kwa bidhaa dot ya kazi mbili vector-thamani, na
3. kwa bidhaa msalaba wa kazi mbili za thamani ya vector.

Theorem: Mali ya Derivative ya Kazi za Vector-Thamani

Hebu\(\vecs{r}\) na\(\vecs{u}\) kuwa differentiable kazi vector-thamani ya\(t\), basi\(f\) kutofautisha halisi thamani ya kazi ya\(t\), na basi\(c\) kuwa scalar.

\[\begin{array}{lrcll} \mathrm{i.} & \dfrac{d}{\,dt}[c\vecs r(t)] & = & c\vecs r′(t) & \text{Scalar multiple} \nonumber\\ \mathrm{ii.} & \dfrac{d}{\,dt}[\vecs r(t)±\vecs u(t)] & = & \vecs r′(t)±\vecs u′(t) & \text{Sum and difference} \nonumber\\ \mathrm{iii.} & \dfrac{d}{\,dt}[f(t)\vecs u(t)] & = & f′(t)\vecs u(t)+f(t)\vecs u′(t) & \text{Scalar product} \nonumber\\ \mathrm{iv.} & \dfrac{d}{\,dt}[\vecs r(t)⋅\vecs u(t)] & = & \vecs r′(t)⋅\vecs u(t)+\vecs r(t)⋅\vecs u′(t) & \text{Dot product} \nonumber\\ \mathrm{v.} & \dfrac{d}{\,dt}[\vecs r(t)×\vecs u(t)] & = & \vecs r′(t)×\vecs u(t)+\vecs r(t)×\vecs u′(t) & \text{Cross product} \nonumber\\ \mathrm{vi.} & \dfrac{d}{\,dt}[\vecs r(f(t))] & = & \vecs r′(f(t))⋅f′(t) & \text{Chain rule} \nonumber\\ \mathrm{vii.} & \text{If} \; \vecs r(t)·\vecs r(t) & = & c, \text{then} \; \vecs r(t)⋅\vecs r′(t) \; =0 \; . & \mathrm{} \nonumber \end{array} \nonumber \]

Ushahidi

Ushahidi wa mali mbili za kwanza hufuata moja kwa moja kutoka kwa ufafanuzi wa derivative ya kazi yenye thamani ya vector. Mali ya tatu inaweza kupatikana kutoka kwa mali mbili za kwanza, pamoja na utawala wa bidhaa. Hebu\(\vecs u(t)=g(t)\,\mathbf{\hat{i}}+h(t)\,\mathbf{\hat{j}}\). Kisha

\ [kuanza {align*}\ dfrac {d} {\, dt} [f (t)\ vecs u (t)] &=\ dfrac {d} {\, dt} [f (t) (g (t)\,\ hatbf {\ kofia {i}} +h (t)\\,\ mathbf {\ kofia}}]\\ [4pt]
&=\ dfrac {d} {\, dt} [f (t) g (t)\,\ hatbf {\ kofia {i}} +f (t) h (t)\,\ hatbf {\ kofia {j}}]\\ [4pt]
&=\ dfrac {d} {\, dt} [f (t) g (t) g (t)]\,\ mathbf {\ kofia {i}} +\ dfrac {d} {\, dt} [f (t) h (t)]\,\ mathbf {\ kofia {j}}\\ [4pt]
&= (f (t) g (t) +f (t) g (t))\,\ hatbf {\ kofia {i}} + (f (t) h (t) +f (t) h (t))\,\ mathbf {\ kofia {j}}\\ [4pt]
&=f (t)\ vecs u (t) +f (t)\ vecs u (t)\ vecs u (t). \ mwisho {align*}\ nonumber\]

Kuthibitisha mali iv. basi\(\vecs r(t)=f_1(t) \,\mathbf{\hat{i}}+g_1(t) \,\mathbf{\hat{j}}\) na\(\vecs u(t)=f_2(t) \,\mathbf{\hat{i}}+g_2(t) \,\mathbf{\hat{j}}\). Kisha

\ [kuanza {align*}\ dfrac {d} {\, dt} [\ vecs r (t)}\ vecs u (t)] &=\ dfrac {d} {\, dt} [f_1 (t) f_2 (t) +g_1 (t) g_2 (t)]\\ [4pt]
&=f_1′ (t) f_1 (t) _2 (t) +f_1 (t) f_2′ (t) +g_1′ (t) g_2 (t) +g_1 (t) g_2′ (t) =f_1′ (t) f_2 (t) +g_1 (t) g_2 (t) +f_1 (t) +g_1 (t) g_2′ (t)\\ [4pt]
& =( f_1′\,\ mathbf {\ kofia {i}} +g_1′\,\ mathbf {\ kofia {j}}) - (f_2\,\ mathbf {i}} +g_2\,\ hatbf {\ kofia {j}}) + (f_1\,\ hatbf {i}}} +g_1\,\ hatbf {\ kofia {j}}) (f_2′\,\ mathbf {j}}) (f_2′\,\ mathbf {j}} thbf {\ kofia {i}} +g_2′\,\ mathbf {\ kofia {j}})\\ [4pt]
&=\ vecs r (t)\ vecs u (t) +\ vecs r (t) +\ vecs r (t)\ vecs u (t). \ mwisho {align*}\ nonumber\]

Ushahidi wa mali v. ni sawa na ile ya mali iv. Mali vi. inaweza kuthibitishwa kwa kutumia utawala wa mnyororo. Mwisho, mali vii. ifuatavyo kutoka mali iv:

\ [kuanza {align*}\ dfrac {d} {\, dt} [\ vecs r (t) ·\ vecs r (t)] &=\ dfrac {d} {\, dt} [c]
\\ [4pt]\ vecs r (t) ·\ vecs r (t) & = 0\\ [4pt]
2\ vecs r (t) ·\ vecs r (t) &= 0\\ [4pt]
\ vecs r (t) ·\ vecs r (t) &= 0\ mwisho {align*}\ nonumber\]

Sasa kwa mifano fulani kutumia mali hizi.

Mfano\(\PageIndex{3}\): Using the Properties of Derivatives of Vector-Valued Functions

Kutokana na kazi ya thamani ya vector

\[\vecs{r}(t)=(6t+8)\,\mathbf{\hat{i}}+(4t^2+2t−3)\,\mathbf{\hat{j}}+5t \,\mathbf{\hat{k}} \nonumber \]

na

\[\vecs{u}(t)=(t^2−3)\,\mathbf{\hat{i}}+(2t+4)\,\mathbf{\hat{j}}+(t^3−3t)\,\mathbf{\hat{k}}, \nonumber \]

tumia kila moja ya derivatives zifuatazo kwa kutumia mali ya derivative ya kazi za thamani ya vector.

1. \(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{u}(t)]\)
2. \(\dfrac{d}{\,dt}[ \vecs{u} (t) \times \vecs{u}′(t)]\)

Suluhisho

Tuna\(\vecs{r}′(t)=6 \,\mathbf{\hat{i}}+(8t+2) \,\mathbf{\hat{j}}+5 \,\mathbf{\hat{k}}\) na\(\vecs{u}′(t)=2t \,\mathbf{\hat{i}}+2 \,\mathbf{\hat{j}}+(3t^2−3) \,\mathbf{\hat{k}}\). Kwa hiyo, kulingana na mali iv:

1. \ [kuanza {align*}\ dfrac {d} {\, dt} [\ vecs r (t)}\ vecs u (t)] &=\ vecs r (t)\\ vecs u (t) +\ vecs r (t)\\ vecs u (t)\\ [4pt]
   &= (6\,\ mathbf {\ hat {}} + (8t+2)\,\ mathbf {\ kofia {j}} +5\,\ hatbf {\ kofia {k}}) (t ^ 2,13)\,\ hatbf {\ kofia {i}} + (2t+4)\,\ hatbf {\ kofia {j}} + (t ^ 3—3t)\,\ mathbf {\ kofia {k}})\\ [4pt]
   &\; + (6t+8)\,\ mathbf {\ kofia {i}} + (4t ^ 2+2t-3)\,\ mathbf {\ kofia {j}} +5t\,\ hatbf {\ kofia {k}}) (2t\,\ mathbf {\ kofia {i}} +2\,\ Mathbf {\ kofia {j}} + (3t ^ 2,13)\,\ Mathbf {\ kofia {k}})\\ [4pt]
   &= 6 (t ^ 2,13) + (8t+2) (2t+4) +5 (t ^ 3,13t)\\ [4pt]
   &\; +2t (6t+8) +2 (4t ^ 2+2t-1) 3) +5t (3t ^ 2,13)\\ [4pt ]
   &= 20t ^ 3+42t ^ 2+26t-16. \ mwisho {align*}\]
2. Kwanza, tunahitaji kukabiliana mali v kwa tatizo hili:

   \[\dfrac{d}{\,dt}[ \vecs{u}(t) \times \vecs{u}′(t)]=\vecs{u}′(t)\times \vecs{u}′(t)+ \vecs{u}(t) \times \vecs{u}′′(t). \nonumber \]

   Kumbuka kwamba bidhaa ya msalaba wa vector yoyote yenyewe ni sifuri. Aidha,\(\vecs u′′(t)\) inawakilisha derivative pili ya\(\vecs u(t):\)

   \[\vecs u′′(t)=\dfrac{d}{\,dt}[\vecs u′(t)]=\dfrac{d}{\,dt}[2t \,\mathbf{\hat{i}}+2 \,\mathbf{\hat{j}}+(3t^2−3) \,\mathbf{\hat{k}}]=2 \,\mathbf{\hat{i}}+6t \,\mathbf{\hat{k}}. \nonumber \]

   Kwa hiyo,

   \ [kuanza {align*}\ dfrac {d} {\, dt} [\ vecs u (t)\ mara\ vecs u (t)] &=0+ (t ^ 2,13)\,\ kofia {\ mathbf {i}} + (2t+4)\,\ kofia {\ mathbf {j}} + (t ^ 3,13t)\,\ kofia {\ mathbf {k}})\ mara (2\,\ kofia {\ mathbf {i}} +6t\,\ kofia {\ mathbf {k}})\\ [4pt]
   &=\ kuanza {vmatrix}\,\ kofia {\ mathbf {i}} &\,\ kofia {\ mathbf {j}} &\,\ kofia {\ mathbf {k}}\\ t ^ 2-3 & 2t+4 & t ^ 3 -3t\\ 2 & 0 & 6t\ mwisho {vMatrix}\\ [4pt]
   & = 6t (2t+4)\,\ kofia {\ hisabati {i}}} - (6t (t ^ 2,1-3) -2 (t ^ 3—3t))\,\ kofia {\ mathbf {j}} -1 (2t+4)\,\ kofia {\ mathbf {k}}\\ [4pt]
   & =( 12t ^ 2+24t)\,\ kofia {\ mathbf {i}} + (12t-4t ^ 3)\,\ kofia {\ mathbf {j}} (4t+8)\, \ kofia {\ mathbf {k}}. \ mwisho {align*}\]

Zoezi\(\PageIndex{3}\)

Tumia\(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{r}′(t)]\) na\( \dfrac{d}{\,dt}[\vecs{u}(t) \times \vecs{r}(t)]\) kwa kazi za thamani ya vector:

• \(\vecs{r}(t)=\cos t \,\mathbf{\hat{i}}+ \sin t \,\mathbf{\hat{j}}−e^{2t} \,\mathbf{\hat{k}}\)
• \(\vecs{u}(t)=t \,\mathbf{\hat{i}}+ \sin t \,\mathbf{\hat{j}}+ \cos t \,\mathbf{\hat{k}}\),

Kidokezo

Fuata hatua sawa na katika Mfano\(\PageIndex{3}\).

Jibu

\(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{r}′(t)]=8e^{4t}\)

\( \dfrac{d}{\,dt}[ \vecs{u}(t) \times \vecs{r}(t)] =−(e^{2t}(\cos t+2 \sin t)+ \cos 2t) \,\mathbf{\hat{i}}+(e^{2t}(2t+1)− \sin 2t) \,\mathbf{\hat{j}}+(t \cos t+ \sin t− \cos 2t) \,\mathbf{\hat{k}}\)

Tangent wadudu na kitengo Tangent wadudu

Kumbuka kwamba derivative katika hatua inaweza kutafsiriwa kama mteremko wa mstari wa tangent kwenye grafu wakati huo. Katika kesi ya kazi yenye thamani ya vector, derivative hutoa vector tangent kwa curve inawakilishwa na kazi. Fikiria kazi yenye thamani ya vector

\[\vecs{r}(t)=\cos t \,\mathbf{\hat{i}} + \sin t \,\mathbf{\hat{j}} \label{eq10} \]

Derivative ya kazi hii ni

\[\vecs{r}′(t)=−\sin t \,\mathbf{\hat{i}} + \cos t \,\mathbf{\hat{j}} \nonumber \]

Ikiwa tunabadilisha thamani\(t=π/6\) katika kazi zote mbili tunapata

\[\vecs{r} \left(\dfrac{π}{6}\right)=\dfrac{\sqrt{3}}{2} \,\mathbf{\hat{i}}+\dfrac{1}{2}\,\mathbf{\hat{j}} \nonumber \]

na

\[ \vecs{r}′ \left(\dfrac{π}{6} \right)=−\dfrac{1}{2}\,\mathbf{\hat{i}}+\dfrac{\sqrt{3}}{2}\,\mathbf{\hat{j}}. \nonumber \]

Grafu ya kazi hii inaonekana kwenye Kielelezo\(\PageIndex{1}\), pamoja na vectors\(\vecs{r}\left(\dfrac{π}{6}\right)\) na\(\vecs{r}' \left(\dfrac{π}{6}\right)\).

Kielelezo\(\PageIndex{1}\): Mstari wa tangent katika hatua ni mahesabu kutoka kwa derivative ya kazi yenye thamani ya vector\(\vecs{r}(t)\).

Angalia kwamba vector\(\vecs{r}′\left(\dfrac{π}{6}\right)\) ni tangent kwa mduara katika hatua sambamba na\(t=\dfrac{π}{6}\). Huu ni mfano wa vector tangent kwa Curve ndege inavyoelezwa na Equation\ ref {eq10}.

Ufafanuzi: kuu kitengo tangent vector

Hebu\(C\) kuwa Curve inavyoelezwa na kazi vector yenye thamani\(\vecs{r}\), na kudhani kwamba\(\vecs{r}′(t)\) ipo wakati\(\mathrm{t=t_0}\) vector tangent\(\vecs{r}\) katika\(t=t_0\) ni vector yoyote kama kwamba, wakati mkia wa vector ni kuwekwa katika hatua\(\vecs r(t_0)\) juu ya grafu, vector\(\vecs{r}\) ni tangent kwa Curve \(C\). Vector\(\vecs{r}′(t_0)\) ni mfano wa vector tangent katika hatua\(t=t_0\). Zaidi ya hayo, kudhani kwamba\(\vecs{r}′(t)≠0\). kuu kitengo tangent vector katika\(t\) ni defined kuwa

\[\vecs{T}(t)=\dfrac{ \vecs{r}′(t)}{‖\vecs{r}′(t)‖}, \nonumber \]

zinazotolewa\(‖\vecs{r}′(t)‖≠0\).

kitengo tangent vector ni nini hasa inaonekana kama: kitengo vector kwamba ni tangent kwa Curve. Ili kuhesabu kitengo cha vector tangent, kwanza pata derivative\(\vecs{r}′(t)\). Pili, mahesabu ya ukubwa wa derivative. Hatua ya tatu ni kugawanya derivative kwa ukubwa wake.

Mfano\(\PageIndex{4}\): Finding a Unit Tangent Vector

Kupata kitengo tangent vector kwa kila moja ya yafuatayo kazi vector-thamani:

1. \(\vecs{r}(t)=\cos t \,\mathbf{\hat{i}}+\sin t \,\mathbf{\hat{j}}\)
2. \(\vecs{u}(t)=(3t^2+2t) \,\mathbf{\hat{i}}+(2−4t^3)\,\mathbf{\hat{j}}+(6t+5)\,\mathbf{\hat{k}}\)

Suluhisho

1. \(\begin{array}{lrcl} \text{First step:} & \vecs r′(t) & = & − \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}} \\ \text{Second step:} & ‖\vecs r′(t)‖ & = & \sqrt{(− \sin t)^2+( \cos t)^2} = 1 \\ \text{Third step:} & \vecs T(t) & = & \dfrac{\vecs r′(t)}{‖\vecs r′(t)‖}=\dfrac{− \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}}}{1}=− \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}} \end{array}\)
2. \(\begin{array}{lrcl} \text{First step:} & \vecs r′(t) & = & (6t+2) \,\hat{\mathbf{i}}−12t^2 \,\hat{\mathbf{j}}+6 \,\hat{\mathbf{k}} \\ \text{Second step:} & ‖\vecs r′(t)‖ & = & \sqrt{(6t+2)^2+(−12t^2)^2+6^2} \\ \text{} & \text{} & = & \sqrt{144t^4+36t^2+24t+40} \\ \text{} & \text{} & = & 2 \sqrt{36t^4+9t^2+6t+10} \\ \text{Third step:} & \vecs T(t) & = & \dfrac{\vecs r′(t)}{‖\vecs r′(t)‖}=\dfrac{(6t+2) \,\hat{\mathbf{i}}−12t^2 \,\hat{\mathbf{j}}+6 \,\hat{\mathbf{k}}}{2 \sqrt{36t^4+9t^2+6t+10}} \\ \text{} & \text{} & = & \dfrac{3t+1}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{i}} - \dfrac{6t^2}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{j}} + \dfrac{3}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{k}} \end{array}\)

Zoezi\(\PageIndex{4}\)

Pata vector kitengo cha tangent kwa kazi yenye thamani ya vector

\[\vecs r(t)=(t^2−3)\,\mathbf{\hat{i}}+(2t+1) \,\mathbf{\hat{j}}+(t−2) \,\mathbf{\hat{k}}. \nonumber \]

Kidokezo

Fuata hatua sawa na katika Mfano\(\PageIndex{4}\).

Jibu

\[\vecs T(t)=\dfrac{2t}{\sqrt{4t^2+5}}\,\mathbf{\hat{i}}+\dfrac{2}{\sqrt{4t^2+5}}\,\mathbf{\hat{j}}+\dfrac{1}{\sqrt{4t^2+5}}\,\mathbf{\hat{k}} \nonumber \]

Integrals ya Vector-Thamani Kazi

Sisi ilianzisha antiderivatives ya kazi halisi ya thamani katika Antiderivatives na integrals uhakika wa kazi halisi ya thamani katika Integral uhakika. Kila moja ya dhana hizi zinaweza kupanuliwa kwa kazi za thamani ya vector. Pia, kama tunaweza kuhesabu derivative ya kazi yenye thamani ya vector kwa kutofautisha kazi za sehemu tofauti, tunaweza kuhesabu antiderivative kwa namna ile ile. Zaidi ya hayo, Theorem ya Msingi ya Calculus inatumika kwa kazi za thamani ya vector pia.

Antiderivative ya kazi yenye thamani ya vector inaonekana katika programu. Kwa mfano, ikiwa kazi yenye thamani ya vector inawakilisha kasi ya kitu wakati t, basi antiderivative yake inawakilisha nafasi. Au, ikiwa kazi inawakilisha kasi ya kitu kwa wakati fulani, basi antiderivative inawakilisha kasi yake.

Ufafanuzi: Integrals ya uhakika na isiyojulikana ya Kazi za Vector-Thamani

Hebu\(f\)\(g\), na\(h\) kuwa integrable halisi ya thamani ya kazi juu ya muda kufungwa\([a,b].\)

1. Muhimu usiojulikana wa kazi yenye thamani ya vector\(\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}\) ni

   \[\int [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}]\,dt= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}}. \nonumber \]

   Muhimu wa uhakika wa kazi yenye thamani ya vector ni

   \[\int_a^b [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}]\,dt = \left[ \int_a^b f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int_a^b g(t)\,dt \right] \,\hat{\mathbf{j}}. \nonumber \]

2. Muhimu usiojulikana wa kazi yenye thamani ya vector\(\vecs r(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}+h(t) \,\hat{\mathbf{k}}\) ni

   \[\int [f(t) \,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int h(t)\,dt \right] \,\hat{\mathbf{k}}. \nonumber \]

   Muhimu wa uhakika wa kazi ya thamani ya vector ni

   \[\int_a^b [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[ \int_a^b f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int_a^b g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int_a^b h(t)\,dt \right] \,\hat{\mathbf{k}}. \nonumber \]

Kwa kuwa muhimu kwa muda usiojulikana wa kazi yenye thamani ya vector inahusisha integrals isiyojulikana ya kazi za sehemu, kila moja ya vipengele hivi vya sehemu ina ushirikiano wa mara kwa mara. Wote wanaweza kuwa tofauti. Kwa mfano, katika kesi mbili-dimensional, tunaweza kuwa na

\[\int f(t)\,dt=F(t)+C_1 \; and \; \int g(t)\,dt=G(t)+C_2, \nonumber \]

wapi\(F\) na\(G\) ni antiderivatives ya\(f\) na\(g\), kwa mtiririko huo. Kisha

\ [kuanza {align*}\ int [f (t)\,\ kofia {\ mathbf {i}} +g (t)\,\ kofia {\ mathbf {j}}]\, dt &=\ kushoto [\ int f (t)\, dt\ haki]\,\ kofia {\ mathbf {i}} +\ kushoto [\ int g (t)\, dt\ haki]\,\ kofia {\ mathbf {j}}\\ [4pt]
&= (F (t) +C_1)\,\ kofia {\ mathbf {i}} + (G (t) +C_2)\,\ kofia {\ mathbf {j}}\\ [4pt] &=F (t)\,\ kofia {\ mathbf {j}\ [4pt] &=F (t)\,\ kofia {\ mathbf {j}\ [4pt]
&=F (t)\,\ kofia {\ mathbf { mathbf {i}} +G (t)\,\ kofia {\ mathbf {j}} +C_1\,\ kofia {\ mathbf {i}} +C_2\,\ kofia {\ mathbf {j}}\\ [4pt]
&= F (t)\,\ kofia {\ mathbf i}} +G (t)\,\ kofia {\ mathbf {j}} +\ vecs {C}\ mwisho {align*}\]

wapi\(\vecs{C}=C_1 \,\hat{\mathbf{i}}+C_2 \,\hat{\mathbf{j}}\). Kwa hiyo, mara kwa mara ya ushirikiano inakuwa vector mara kwa mara.

Mfano\(\PageIndex{5}\): Integrating Vector-Valued Functions

Tumia kila moja ya vipengele vifuatavyo:

1. \( \displaystyle \int [(3t^2+2t) \,\hat{\mathbf{i}}+(3t−6) \,\hat{\mathbf{j}}+(6t^3+5t^2−4) \,\hat{\mathbf{k}}]\,dt\)
2. \( \displaystyle \int [⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩] \,dt\)
3. \( \displaystyle \int_{0}^{\frac{\pi}{3}} [\sin 2t \,\hat{\mathbf{i}}+ \tan t \,\hat{\mathbf{j}}+e^{−2t} \,\hat{\mathbf{k}}]\,dt\)

Suluhisho

1. Tunatumia sehemu ya kwanza ya ufafanuzi wa sehemu muhimu ya pembe ya nafasi:
2. \ [kuanza {align*}\ int [(3t^2+2t)\,\ kofia {\ mathbf {i}} + (3t--6)\,\ kofia {\ mathbf {j}} + (6t ^ 3+5t ^ 2,14)\,\ kofia {\ mathbf {k}}]\, dt &=\ kushoto [\ int 3,14 t ^ 2+2t\, dt\ haki]\,\ kofia {\ mathbf {i}} +\ kushoto [\ int 3t-6\, dt\ haki]\,\ kofia {\ mathbf {j}} +\ kushoto [\ int 6t ^ 3+5t ^ 2—4\, dt\ haki]\,\ kofia {\ mathbf {k}}\ [4pt]
   & =( t ^ 3+t ^ 2) \,\ kofia {\ mathbf {i}} +\ kushoto (\ frac {3} {2} t ^ 2,16t\ haki)\,\ kofia {\ mathbf {j}} +\ kushoto (\ frac {3} {2} t^4+\ frac {5} {3} t ^ 3,14t\ haki)\,\ kofia {\ mathbf {3}\ haki)\,\ kofia {\ mathbf {3} f {k}} +\ vecs C.\ mwisho {align*}\]
3. Kwanza hesabu\(⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩:\)

   \ [kuanza {align*} t^t, t ^ 2, t ^ 3\ nyakati t^3, t ^ 2, t&=\ kuanza {vMatrix}\ kofia {\ matriki}\ hath {i}} &\,\ kofia {\ matriki}\\ t & t ^ 2 & t ^ 3\\ t ^ 3 & t ^ 2 & t\ mwisho {vmatrix}\ [4pt]
   &= (t ^ 2 (t) -t ^ 3 (t ^ 2))\,\ kofia {\ mathbf {i}}} (t ^ 2,1t ^ 3 (t ^ 3))\,\ kofia {\ mathbf {j}} + (t ^ 2) -t ^2 (t ^ 3))\,\ kofia {\ mathbf {k}}\\ [4pt]
   & =( t ^ 3—t ^ 5)\,\ kofia {\ mathbf {i}} + (t ^ 6,1t ^ 2)\,\ kofia {\ mathbf {j}} + (t ^ 3,1t^5)\,\ kofia {\ mathbf {j}} + (t ^ 3,1t^5)\,\ kofia {\ mathbf {j}} + (t ^ 3,1t^5)\,\ kofia {\ mathbf {j}} thbf {k}}. \ mwisho {align*}\ nonumber\]

   Kisha, badala hii nyuma katika muhimu na kuunganisha:

   \ [kuanza {align*}\ int [t^t, t^2, t ^ 3\ nyakati t^3, t ^ 2, t]\, dt &=\ int (t ^ 3,1t ^ 5)\,\ kofia {\ mathbf {i}} + (t ^ 6—t ^ 2)\,\ kofia {\ mathbf {j}} + (t ^ 6- t ^ 2)\,\ kofia {\ mathbf {j}} + (t ^ 6- t ^ 2)\,\ kofia {\ mathbf {j}} + (t ^ 6- t ^ 2)\,\ kof3,1t^5)\,\ kofia {\ mathbf {k}}\, dt\\ [4pt]
   &=\ kushoto (\ frac {t ^ 4} {4} -\ frac {t ^ 6} {6}\ haki)\,\ kofia {\ mathbf {i}} +\ kushoto (\ frac {t ^ 7}}}}\ frac {7}}}\ Frac {7}}}\ frac {t ^ 3} {3}\ haki)\,\ kofia {\ mathbf {j}} +\ kushoto (\ frac {t ^ 4} {4} -\ frac {t ^ 6} {6}\ haki)\,\ kofia {\ mathbf {k}} +\ vecs C.\ mwisho {align*}\]

4. Tumia sehemu ya pili ya ufafanuzi wa sehemu muhimu ya pembe ya nafasi:

   \ [kuanza {align*}\ int_0^ {\ frac {\ pi} {3}} [\ sin 2t\,\ kofia {\ mathbf {i}} +\ tan t\,\ kofia {\ mathbf {j}} +e^ {-2t}\,\ kofia {\ mathbf {k}}]\, dt &=\ kushoto [\ int_0^ {\ frac {π} {3}}\ sin 2t\, dt\ haki]\,\ kofia {\ mathbf {i}} +\ kushoto [\ int_0^ {\ frac {π} {3}}\ tan t\, dt\ haki]\,\ kofia {\ mathbf {j}} +\ kushoto [\ int_0^ {\ int_0^ {\ frac {\ int_0^ {\ frac {\ int_0^ {\ frac {\ int_0^ {\ frac {\ int_0^ {\ frac {\ int_c {π} {3}} e^ {-2t}\, dt\ haki] \,\ kofia {\ mathbf {k}}\\ [4pt]
   &= (-\ tfrac {1} {2}\ cos 2t)\ Big\ vert_ {0} ^ {π/3}\,\ kofia {\ mathbf {i}}} (\ ln |\ cos t|)\ Big\ vert_ {0} ^ {π/3}\,\ kofia {\ mathbf {j}}} -\ kushoto (\ tfrac {1} {2} e^ {-2t}\ haki)\ Big\ vert_ {0} ^ {π/3}\,\ kofia {\ mathbf {k}}\\ [4pt]
   &=\ kushoto (\\ tfrac {1}}\ cos\ tfrac {1}\ cos\ tfrac {2π} {3} +\ tfrac {1} {2}\ cos 0\ kulia)\,\ kofia {\ mathbf {i}}} -\ kushoto (\ ln\ kushoto (\ cos\ tfrac {π} {3}\ kulia) -\ ln (\ cos 0)\ haki)\,\ kofia {\ mathbf {j}}}}}}}\ kushoto (\ tfrac {1} {2} e^ {21-2π/3} -\ tfrac {1} {2} e^ {ї 2 (0)}\ haki)\,\ kofia {\ mathbf {k}}\\ [4pt]
   & =\ kushoto (\ tfrac {1} {4} +\ tfrac {1} {1} {2}\ haki)\,\ kofia {\ mathbf {i}}} (-\ ln 2)\,\ kofia {\ mathbf {j}} -\ kushoto (\ tfrac {1} {2} e^ {-2π/3}} -\ tfrac {1} {2}\ haki)\,\ kofia {\ mathbf {k}}\ [4pt]
   &=\ tfrac {3} {4}\,\ kofia {\ mathbf {i}} (+\ tfrac {3} {4}\,\ kofia {\ mathbf {i}} ln 2)\,\ kofia {\ mathbf {j}} +\ kushoto (\ tfrac {1} {2}}}\ tfrac {1} {2} e^ {-2π/3}\ haki)\,\ kofia {\ mathbf {k}}. \ mwisho {align*}\]

Zoezi\(\PageIndex{5}\)

Tumia muhimu yafuatayo:

\[\int_1^3 [(2t+4) \,\mathbf{\hat{i}}+(3t^2−4t) \,\mathbf{\hat{j}}]\,dt \nonumber \]

Kidokezo

Tumia ufafanuzi wa umuhimu wa uhakika wa safu ya ndege.

Jibu

\[\int_1^3 [(2t+4) \,\mathbf{\hat{i}}+(3t^2−4t) \,\mathbf{\hat{j}}]\,dt = 16 \,\mathbf{\hat{i}}+10 \,\mathbf{\hat{j}} \nonumber \]

Muhtasari

• Ili kuhesabu derivative ya kazi yenye thamani ya vector, kuhesabu derivatives ya kazi za sehemu, kisha uwaweke tena kwenye kazi mpya ya thamani ya vector.
• Mali nyingi za kutofautisha kazi za scalar pia zinatumika kwa kazi za thamani ya vector.
• Derivative ya kazi yenye thamani ya vector pia\(\vecs r(t)\) ni vector tangent kwa curve. Kitengo cha tangent vector\(\vecs T(t)\) kinahesabiwa kwa kugawanya derivative ya kazi yenye thamani ya vector kwa ukubwa wake.
• Antiderivative ya kazi yenye thamani ya vector hupatikana kwa kutafuta antiderivatives ya kazi za sehemu, kisha kuwaweka pamoja katika kazi yenye thamani ya vector.
• Muhimu wa uhakika wa kazi yenye thamani ya vector hupatikana kwa kutafuta viungo vya uhakika vya kazi za sehemu, kisha kuwaweka pamoja katika kazi yenye thamani ya vector.

Mlinganyo muhimu

• Derivative ya kazi yenye thamani ya vector\[\vecs r′(t) = \lim \limits_{\Delta t \to 0} \dfrac{\vecs r(t+\Delta t)−\vecs r(t)}{ \Delta t} \nonumber \]
• Kitengo kuu tangent vector\[\vecs T(t)=\frac{\vecs r′(t)}{‖\vecs r′(t)‖} \nonumber \]
• Muda usiojulikana muhimu wa kazi yenye thamani ya vector\[\int [f(t) \,\mathbf{\hat{i}}+g(t)\,\mathbf{\hat{j}} + h(t) \,\mathbf{\hat{k}}]\,dt= \left[ \int f(t)\,dt \right] \,\mathbf{\hat{i}}+ \left[ \int g(t)\,dt \right] \,\mathbf{\hat{j}} + \left[ \int h(t)\,dt \right] \,\mathbf{\hat{k}}\nonumber \]
• Ufafanuzi muhimu wa kazi yenye thamani ya vector\[\int_a^b [f(t) \,\mathbf{\hat{i}}+g(t) \,\mathbf{\hat{j}} + h(t) \,\mathbf{\hat{k}}]\,dt= \left[\int_a^b f(t)\,dt \right] \,\mathbf{\hat{i}}+ \left[ \int _a^b g(t)\,dt \right] \,\mathbf{\hat{j}} + \left[ \int _a^b h(t)\,dt \right] \,\mathbf{\hat{k}}\nonumber \]

faharasa

muhimu muhimu ya kazi yenye thamani ya vector

vector iliyopatikana kwa kuhesabu muhimu ya kila sehemu ya kazi ya kazi iliyotolewa yenye thamani ya vector, kisha kutumia matokeo kama vipengele vya kazi inayosababisha

derivative ya kazi yenye thamani ya vector

derivative ya kazi vector-thamani\(\vecs{r}(t)\) ni\(\vecs{r}′(t) = \lim \limits_{\Delta t \to 0} \frac{\vecs r(t+\Delta t)−\vecs r(t)}{ \Delta t}\), mradi kikomo ipo

muda usiojulikana muhimu wa kazi yenye thamani ya vector

kazi yenye thamani ya vector na derivative ambayo ni sawa na kazi iliyotolewa yenye thamani ya vector

kuu kitengo tangent vector

kitengo vector tangent kwa Curve C

vector tangent

kwa\(\vecs{r}(t)\) vector\(t=t_0\) yoyote\(\vecs v\) kama kwamba, wakati mkia wa vector umewekwa kwenye hatua\(\vecs r(t_0)\) kwenye grafu, vector\(\vecs{v}\) ni tangent kwa Curve C