- Differential Equation:
- $\dfrac{dr}{dt} = -ti – tj -tk $
- Initial Condition:
- $ r(0) = i + 2j +3k$
This problem aims to find the initial value of a vector function in the form of a differential equation. For this problem, one needs to understand the concept of initial values, Laplace Transform, and solve differential equations given the initial conditions.
An initial value problem, in multivariable calculus, is defined as a standard differential equation given with an initial condition that defines the value of the unknown function at a given point in a certain domain.
Now coming onto the Laplace transform, which is named after its creator Pierre Laplace, is an integral transform that transforms an arbitrary function of a real variable into a function of a complex variable $s$.
Expert Answer:
Here, we have a simple first-order derivative and some initial conditions, so first we will be required to find a precise solution to this problem. One thing to note here is that the only condition we have will let us solve for the one constant we select when we integrate.
As we have defined above that if any problem is given to us as a derivative and with initial conditions to solve for an explicit solution is known as an initial value problem.
So we will start first by taking the differential equation and rearranging it for the value of $r$:
\[dr = (-ti – tj -tk) dt \]
Integrating on both sides:
\[ \int dr = \int(-ti – tj -tk) dt \]
Solving the Integral:
\[ r(t) = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + C \]
Putting the initial condition here $r(0)$:
\[ r(0) = 0i – 0j – 0k + C \]
One expression of $r(0)$ is given in question so we’re going to put both the expressions of $r(0)$ as equals:
\[ 0i – 0j – 0k + C = i + 2j +3k \]
$C$ comes out to be:
\[ C = i + 2j +3k \]
Now plugging $C$ back in $r$:
\[ r = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + C\]
\[ r = – \dfrac{t^2}{2}i – \dfrac{t^2}{2}j – \dfrac{t^2}{2}k + i + 2j +3k \]
Numerical Result:
\[ r = – \left( \dfrac{t^2}{2} + 1\right)i – \left(\dfrac{t^2}{2}+2 \right)j – \left(\dfrac{t^2}{2}+3\right)k \]
Example:
Solve the initial value problem for $r$ as a vector function of $t$.
Differential Equation:
\[\dfrac{dr}{dt} = -3ti – 3tj -tk \]
Initial Condition:
\[ r(0) = 2i + 4j +9k\]
Rearranging for $r$:
\[dr = (-3ti – 3tj -tk) dt \]
Integrating on both sides:
\[\int dr = \int(-3ti -3tj -tk) dt \]
Solving the Integral:
\[r = – \dfrac{-3t^2}{2}i – \dfrac{-3t^2}{2}j – \dfrac{t^2}{2}k + C \]
Putting $r(0)$:
\[ r(0) = 0i – 0j – 0k + C \]
Putting both expressions of $r(0) equals:$
\[ 0i – 0j – 0k + C = 2i + 4j +9k\]
$C$ comes out to be:
\[ C = 2i + 4j +9k \]
Now plugging $C$ back in $r$:
\[ r = – \dfrac{-3t^2}{2}i – \dfrac{-3t^2}{2}j – \dfrac{t^2}{2}k + 2i + 4j +9k \]
\[ r = \left( 2 – \dfrac{3t^2}{2}\right)i + \left( 4 -\dfrac{3t^2}{2} \right)j + \left(9 – \dfrac{t^2}{2}\right)k \]