Solve the initial value problems in Exercises 11-20 for 𝐫 as a vector function of t . Diffe

Solve the initial value problems in Exercises 11-20 for 𝐫 as...

Solve the initial value problems in Exercises $11-20$ for $\mathbf{r}$ as a vector function of $t .$ $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=100 \mathbf{j}}\end{array} $$

Solve the initial value problems in Exercises $11-20$ for $\mathbf{r}$ as a vector function of $t .$

$$
\begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=100 \mathbf{j}}\end{array}
$$

Key Concepts

Antiderivatives

Antiderivatives, or indefinite integrals, reverse the process of differentiation. In the context of vector functions, the antiderivative is obtained by integrating each component of the derivative, which results in a general solution that can be refined using initial conditions.

Initial Value Problems

An initial value problem provides the value of the unknown function at a specific point in time. This information is used to determine the constants of integration obtained during the integration process, ensuring that the final solution satisfies the given starting condition.

Vector Differential Equations

A vector differential equation specifies the derivative of a vector function. Instead of a single unknown function, the unknown is a vector whose components are differentiated individually. This allows for solving for each scalar component using standard techniques from differential equations.

Component-wise Integration

Vector functions are integrated by treating each component separately. This means that to find the original vector function, you take the antiderivative of each component of the derivative vector, which simplifies the problem into a set of scalar integrals.

Transcript

00:01 We want to solve the initial value problem that we are given here on the board.

00:06 So the first thing that i'm going to do is i like to work with everything using the bracket notation for vectors.

00:12 So i'm just going to rewrite it like that to start.

00:15 So this first one.

00:16 So remember i and j are going to be the x and y components respectively.

00:21 So we can rewrite this as 180t, then 180t minus 16t.

00:32 Squared.

00:34 And then our initial condition, we can write this as, well, there's no i, so that means i is zero.

00:39 And then our j is just going to be 100.

00:43 So we can rewrite it like this first.

00:47 Now, what we're going to do is go ahead and integrate the differential that we're given.

00:54 And that's going to tell us what r of t is.

00:57 So we have r, not f, but r of t is going to be equal to the integral of 108 t, then 108 minus 16 t squared d t.

01:17 And then remember, we can go ahead and distribute the integral across.

01:23 And so doing this will give, so i'm going to factor that 180 out.

01:28 So 180 t d t, and then distributed across that plus or minus there.

01:35 And we end up with 180 t d t minus 16 integral t squared d t so we'll end up with 180 and then that's going to be t squared over two and then we need to add some constant i'll just call it a and then over here we're going to do the same thing so it's going to be 180 over t squared two minus and then 16 t cubed over to 3 and then plus some constant b now we know that our initial condition are 0 well this here is supposed to be equal to 0 100 so now we can go ahead and set these two things equal to each other to solve for what a and b should end up being right so let's go ahead and do that really quickly...

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