# Determine r(t) if r(t) =7t i + 3 \sqrt{t} j + 7t^2 k and r(1) = 5 i + 4 j + 1k.

## Question:

Determine r(t) if {eq}r'(t) =7t i + 3 \sqrt{t} j + 7t^2 k {/eq} and r(1) = 5 i + 4 j + 1k.

## Integration:

The integration of a differential function of a vector gives the original term of the function. Let {eq}r'\left( x \right) = a{x^n} + bx + c {/eq} be any differential function of a vector. The integration of the given function can be written as follows,

{eq}\int {r'\left( x \right)} = \int {\left( {a{x^n} + m} \right)dx} + \int {\left( {bx + y} \right)dx} + \int {\left( {c + z} \right)dx} {/eq} .

Here, m, y, and z are the integration constant.

## Answer and Explanation:

**Given**

- The derivative vector is given by,

{eq}r'\left( t \right) = 7ti + 3\sqrt t j + 7{t^2}k {/eq}.

- The value of the vector at t=1 is,

{eq}r\left( 1 \right) = 5i + 4j + 1k {/eq}.

Find the vector r(t).

Consider the given derivative.

{eq}r'\left( t \right) = 7ti + 3\sqrt t j + 7{t^2}k {/eq}

Integrate the above vector,

{eq}\begin{align*} \int {r'\left( t \right)} &= \int {\left( {7ti + 3\sqrt t j + 7{t^2}k} \right)dt} \\ & = \left( {\int {7tdt} } \right)i + \left( {\int {3\sqrt t } dt} \right)j + \left( {\int {7{t^2}} dt} \right)k\\ r\left( t \right)& = \left( {\dfrac{7}{2}{t^2}i + 3\left[ {\dfrac{{{t^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right]j + 7\left( {\dfrac{{{t^{2 + 1}}}}{{2 + 1}}} \right)k} \right)\\ & = \left( {\left( {\dfrac{7}{2}{t^2} + {C_1}} \right)i + \left( {2\left[ {{t^{\dfrac{3}{2}}}} \right] + {C_2}} \right)j + \left( {\dfrac{7}{3}\left( {{t^3}} \right) + {C_3}} \right)k} \right) \end{align*} {/eq}

Here, {eq}{C_1}, {C_2} {/eq}, and {eq}{C_3} {/eq} are the integration constant.

Since, {eq}r\left( 1 \right) = 5i + 4j + 1k {/eq}, then at {eq}t=1 {/eq},

{eq}\begin{align*} \left( {\left( {\dfrac{7}{2}{{\left( 1 \right)}^2} + {C_1}} \right)i + \left( {2\left[ {{1^{\dfrac{3}{2}}}} \right] + {C_2}} \right)j + \left( {\dfrac{7}{3}\left( {{1^3}} \right) + {C_3}} \right)k} \right) &= 5i + 4j + 1k\\ \left( {\left( {\dfrac{7}{2} + {C_1}} \right)i + \left( {2 + {C_2}} \right)j + \left( {\dfrac{7}{3} + {C_3}} \right)k} \right) &= 5i + 4j + 1k \end{align*} {/eq}

Compare the components of the I, j, and k.

{eq}\begin{align*} \dfrac{7}{2} + {C_1} &= 5\\ {C_1} &= 5 - \dfrac{7}{2}\\ &= \dfrac{3}{2} \end{align*} {/eq}

Similarly,

{eq}\begin{align*} 2 + {C_2} &= 4\\ {C_2} &= 2 \end{align*} {/eq}

And

{eq}\begin{align*} \dfrac{7}{3} + {C_3} &= 1\\ {C_3} &= 1 - \dfrac{7}{3}\\ & = \dfrac{{ - 4}}{3} \end{align*} {/eq}

Therefore, the vector is,

{eq}r\left( t \right) = \left( {\left( {\dfrac{7}{2}{t^2} + \dfrac{3}{2}} \right)i + \left( {2\left[ {{t^{\dfrac{3}{2}}}} \right] + 2} \right)j + \left( {\dfrac{7}{3}\left( {{t^3}} \right) - \dfrac{4}{3}} \right)k} \right). {/eq}

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13