# Find the function r that satisfies the following conditions. r'(t) = {t^5} / {t^6 + 1} i + t^5...

## Question:

Find the function r that satisfies the following conditions.

{eq}\displaystyle r'(t) = \frac {t^5} {t^6 + 1}\mathbf i + t^5 e^{-t^6}\mathbf j - \frac {3t^2} {\sqrt {t^3 + 9}}\mathbf k\ ;\ r (0) = 6\mathbf i + \frac {23} 6\mathbf j - 2\mathbf k {/eq}.

## Initial Value Problem:

The initial value problem is given with a differential equation along with an initial condition. To solve this, first, we solve the differential equation (mostly by taking the antiderivative on both sides) and then apply the initial condition to solve for the arbitrary constant.

## Answer and Explanation: 1

The given derivative of the function is:

$$r^{\prime}(t)=\frac{t^{5}}{t^{6}+1} \mathbf{i}+t^{5} e^{-t^{6}} \mathbf{j}-\frac{3 t^{2}}{\sqrt{t^{3}+9}} \mathbf{k}$$

Taking the antiderivative on both sides:

$$r(t) =\mathbf{i} \int \frac{t^{5}}{t^{6}+1} dt+\mathbf{j} \int t^{5} e^{-t^{6}} dt - \mathbf{k} \int\frac{3 t^{2}}{\sqrt{t^{3}+9}} dt \,\, \rightarrow (1)$$

Now we evaluate each of these integrals using the substitution method and finally, we would substitute all of them in (1):

First Integral:

\int \frac{t^{5}}{t^{6}+1} dt \\ \text{Let } t^6+1=u \\ \text{Then } 6t^5 dt = du \\ t^5 dt = \dfrac{1}{6} du \text{(Then the above integral becomes)} \\ \begin{align} \int \dfrac{1}{u} \left( \dfrac{1}{6} du \right) & = \dfrac{1}{6} \int \dfrac{1}{u} du \\ &=\dfrac{1}{6} \ln |u| \\ &= \dfrac{1}{6} \ln |t^6+1| \end{align}

Second integral:

\int t^{5} e^{-t^{6}} dt \\ \text{Let } -t^6 = u \\ \text{Then } -6t^5 dt = du \\ t^5 dt = \dfrac{-1}{6} du\\ \text{(Then the above integral becomes)} \\ \begin{align} \int e^u \left( \dfrac{-1}{6} du \right) &= \dfrac{-1}{6} \int e^u du \\ & = \dfrac{-1}{6} e^u \\ &= \dfrac{-1}{6} e^{-t^6} \end{align}

Third integral:

\int\frac{3 t^{2}}{\sqrt{t^{3}+9}} dt \\ \text{Let } t^3+9=u \\ \text{Then } 3t^2 dt = du \\ \text{(Then the above integral becomes)} \\ \begin{align} \int \dfrac{1}{ \sqrt{u}} du & = \int u^{-1/2} du \\ &= \cdot \dfrac{u^{1/2}}{(1/2)} \\ &= 2 \sqrt{t^3+9} \end{align}

Substitute all these values in (1) and we add an integration constant C:

$$r(t) =\dfrac{1}{6} \ln |t^6+1| \mathbf{i} - \dfrac{1}{6} e^{-t^6}\mathbf{j} - 2 \sqrt{t^3+9}\mathbf{k} +C \,\, \rightarrow (2)$$

The given initial condition is:

$$r(0)=6 \mathbf{i}+\frac{23}{6} \mathbf{j}-2 \mathbf{k}$$

Substitute this in (2):

$$6 \mathbf{i}+\frac{23}{6} \mathbf{j}-2 \mathbf{k} = \dfrac{1}{6} \ln 1 \mathbf{i} - \dfrac{1}{6} (1) \mathbf{j} -6\mathbf{k} +C \\ C = (6-0) \mathbf{i}+ \left( \frac{23}{6} + \dfrac{1}{6}\right) \mathbf{j}+ (-2+6) \mathbf{k} = 6 \mathbf{i} + 4 \mathbf{j} + 4 \mathbf{k}$$

Substitute this back in (2):

\begin{align} r(t) &= \dfrac{1}{6} \ln |t^6+1| \mathbf{i} - \dfrac{1}{6} e^{-t^6}\mathbf{j} - 2 \sqrt{t^3+9}\mathbf{k} + ( 6 \mathbf{i} + 4 \mathbf{j} + 4 \mathbf{k}) \\ &= \mathbf{i} \left( \dfrac{1}{6} \ln |t^6+1| + 6\right) + \mathbf{j} \left( - \dfrac{1}{6} e^{-t^6}+4 \right) +\mathbf{k} ( - 2 \sqrt{t^3+9} +4) \end{align}

Therefore: {eq}\boxed{\mathbf{ \mathbf{i} \left( \dfrac{1}{6} \ln |t^6+1| + 6\right) + \mathbf{j} \left( - \dfrac{1}{6} e^{-t^6}+4 \right) +\mathbf{k} ( - 2 \sqrt{t^3+9} +4)}} {/eq}

#### Learn more about this topic:

Initial Value in Calculus: Definition, Method & Example

from

Chapter 11 / Lesson 13
15K

Learn to define the initial value problem and initial value formula. Learn how to solve initial value problems in calculus. See examples of initial value problems.