Find r(t) if r'(t) = t^7 \ i + e^t \ j + 2te^{2t}k and r(0) = i+j+k.


Find {eq}r(t) {/eq} if {eq}r'(t) = t^7 \ i + e^t \ j + 2te^{2t}k {/eq} and {eq}r(0) = i+j+k. {/eq}

Differentiation in Calculus:

This problem is an initial value problem. To solve this problem, we'll integrate both sides of the given function {eq}r(t) {/eq}. In order to do that, we'll use integration by parts.

Next, plug in the given initial value to get the value of the constant.

Answer and Explanation:

We are given:

{eq}\displaystyle r'(t) = t^7 i + e^t j + 2te^{2t} k {/eq}

Integrate both sides:

{eq}\displaystyle \Rightarrow r(t) = \left( \int t^7 \ dt \right) i + \left( \int e^t \ dt \right) j+ \left( \int 2 te^{2t} \ dt \right) k {/eq}

Now solve: {eq}\displaystyle \int t^7 \ dt = \dfrac{t^8}{8} +c_{1} {/eq}

Next solve: {eq}\displaystyle \int e^t \ dt = e^{t}+c_{2} {/eq}

Next solve using integration by parts: {eq}\displaystyle 2 \int te^{2t} \ dt = 2 t \int e^{2t} \ dt - 2 \int \left( (t)' (\int e^{2t} \ dt) \right) \ dt = t e^{2t}- \dfrac{e^{2t}}{2} +c_{3} {/eq}

Plug in back:

{eq}\displaystyle \Rightarrow r(t) = \left( \dfrac{t^8}{8} +c_{1}\right) i +\left( e^t +c_{2} \right) j+ \left( te^{2t} - \dfrac{e^{2t}}{2}+c_{3} \right) k {/eq}

Compute the given condition {eq}r(0) = i + j + k {/eq}

{eq}\displaystyle c_{1} = 1 , c_{2} = 1-1=0 , c_{3}= 1 + \dfrac{1}{2} = \dfrac{3}{2} {/eq}

Plug in:

{eq}\displaystyle \Rightarrow r(t) = \left( \dfrac{t^8}{8} +1 \right) i +\left( e^t \right) j+ \left(te^{2t} - \dfrac{e^{2t}}{2}+ \dfrac{3}{2} \right) k {/eq}

Learn more about this topic:

Differential Calculus: Definition & Applications

from Calculus: Help and Review

Chapter 13 / Lesson 6

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