Solve the initial value problem for {r} as a vector function of t . Differential equation: ...

Question:

Solve the initial value problem for {eq}\mathbf{r} {/eq} as a vector function of {eq}t {/eq}.

Differential equation: {eq}\frac{dr}{dt} = -3t \mathbf{i} - 3t \mathbf{j} - 3t \mathbf{k} {/eq}

Initial condition: {eq}\mathbf{r}(0) = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} {/eq}

Power Rule for Integration

The power rule is used to get the integral of a function {eq}f(t) {/eq} where the variable is raised to a constant real number {eq}n {/eq}. The rule is defined by the following equation:

{eq}\displaystyle \int t^n \;dt = \frac{t^{n+1}}{n+1} \\ {/eq}

Another rule is the sum/difference rule which states that the integral of the sum (or difference) of two functions is the sum (or difference) of the integral of the functions. In equation, this is expressed as:

{eq}\displaystyle \int (f(t) \pm g(t)) \;dt = \int f(t)\;dt \pm \int g(t)\;dt {/eq}

Answer and Explanation:

To get the vector, we will integrate the given differential equation.

{eq}\dfrac{dr}{dt} = -3t \mathbf{i} - 3t \mathbf{j} - 3t \mathbf{k} \\ dr = \left (-3t \mathbf{i} - 3t \mathbf{j} - 3t \mathbf{k} \right ) dt \\ \displaystyle \int dr = \int \left (-3t \mathbf{i} - 3t \mathbf{j} - 3t \mathbf{k} \right ) dt \\ \displaystyle \mathbf{r} = \int \left (-3t \mathbf{i} - 3t \mathbf{j} - 3t \mathbf{k} \right ) dt {/eq}

Apply sum/difference rule: {eq}\displaystyle \int (f(t) \pm g(t)) \;dt = \int f(t)\;dt \pm \int g(t)\;dt {/eq}.

{eq}\displaystyle \mathbf{r} = \int -3t \mathbf{i}\;dt - \int 3t \mathbf{j}\;dt - \int 3t \mathbf{k} dt {/eq}

Take out the constants.

{eq}\displaystyle \mathbf{r} = -3\int t \mathbf{i}\;dt - 3\int t \mathbf{j}\;dt - 3\int t \mathbf{k} dt {/eq}

Apply power rule: {eq}\displaystyle \int t^n \;dt = \frac{t^{n+1}}{n+1} {/eq}.

{eq}\displaystyle \mathbf{r} = -3\left ( \frac{t^{1+1}}{1+1} \right ) + c_1 \mathbf{i} -3\left ( \frac{t^{1+1}}{1+1} \right ) + c_2 \mathbf{j} -3\left ( \frac{t^{1+1}}{1+1} \right ) + c_3 \mathbf{k} {/eq}

where {eq}c_1, c_2 \ and \ c_3 {/eq} are integration constants.

{eq}\displaystyle \mathbf{r}(t) = -3\left ( \frac{t^{2}}{2} \right ) +c_1 \mathbf{i} -3\left ( \frac{t^{2}}{2} \right ) + c_2 \mathbf{j} -3\left ( \frac{t^{2}}{2} \right ) + c_3 \mathbf{k} \\ \displaystyle \mathbf{r}(t) = \left (-\frac{3t^{2}}{2} +c_1 \right ) \mathbf{i} +\left ( -\frac{3t^{2}}{2} + c_2 \right ) \mathbf{j} + \left ( -\frac{3t^{2}}{2} + c_3 \right )\mathbf{k} {/eq}

To get the values of the integration constants, we will use the initial condition given.

{eq}\displaystyle \mathbf{r}(0) = \left (-\frac{3(0)^{2}}{2} +c_1 \right ) \mathbf{i} +\left ( -\frac{3(0)^{2}}{2} + c_2 \right ) \mathbf{j} + \left ( -\frac{3(0)^{2}}{2} + c_3 \right )\mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \\ \displaystyle \left (-\frac{3(0)}{2} +c_1 \right ) \mathbf{i} +\left ( -\frac{3(0)}{2} + c_2 \right ) \mathbf{j} + \left ( -\frac{3(0)}{2} + c_3 \right )\mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \\ \displaystyle \left (-\frac{0}{2} +c_1 \right ) \mathbf{i} +\left ( -\frac{0}{2} + c_2 \right ) \mathbf{j} + \left ( -\frac{0}{2} + c_3 \right )\mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \\ \displaystyle \left (-0 +c_1 \right ) \mathbf{i} +\left ( -0 + c_2 \right ) \mathbf{j} + \left ( -0 + c_3 \right )\mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \\ \displaystyle \left (c_1 \right ) \mathbf{i} +\left ( c_2 \right ) \mathbf{j} + \left (c_3 \right )\mathbf{k} = 2 \mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \\ {/eq}

Equate the components.

{eq}c_1 = 2 \\ c_2 = 4 \\ c_3 = 3 {/eq}

The antiderivative of the given vector is

{eq}\displaystyle \boldsymbol{\mathbf{r}(t) = \left (-\frac{3t^{2}}{2} + 2 \right ) \mathbf{i} +\left ( -\frac{3t^{2}}{2} + 4 \right ) \mathbf{j} + \left ( -\frac{3t^{2}}{2} + 3 \right )\mathbf{k} } {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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