[tex]\mathbf r'(t)=\langle\sin9t,\sin6t,9t\rangle[/tex]
[tex]\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt[/tex]
[tex]\mathbf r(t)=\left\langle\displaystyle\int\sin9t\,\mathrm dt,\int\sin6t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle[/tex]
[tex]\displaystyle\int\sin9t\,\mathrm dt=\frac19\cos9t+C_1[/tex]
[tex]\displaystyle\int\sin6t\,\mathrm dt=\frac16\cos6t+C_2[/tex]
[tex]\displaystyle\int9t\,\mathrm dt=\frac92t^2+C_3[/tex]
With the initial condition [tex]\mathbf r(0)=\langle4,6,3\rangle[/tex], we find
[tex]\dfrac19\cos0+C_1=4\implies C_1=\dfrac{35}9[/tex]
[tex]\dfrac16\cos0+C_2=6\implies C_2=\dfrac{35}6[/tex]
[tex]\dfrac92\cdot0^2+C_3=3\implies C_3=3[/tex]
So the particular solution to the IVP is
[tex]\mathbf r(t)=\left\langle\dfrac19\cos9t+\dfrac{35}9,\dfrac16\cos6t+\dfrac{35}6,\dfrac92t^2+3\right\rangle[/tex]
Here we want to solve a differential equation:
r'(t) = < sin(9t), sin(6t), 9t >
With the restriction: r(0)=⟨4,6,3⟩
Remember that:
r'(t) = dr(t)/dt
Also, remember the general rules:
[tex]f(x) = A\cdot x^n + C\\\\f'(x) = n\cdot A \cdot x^{n - 1}[/tex]
and:
[tex]g(x) = A \cdot cos(k \cdot x) + C \\\\g'(x) = -k \cdot A \cdot sin(k \cdot x)[/tex]
Then using these rules we can get the antiderivatives of r(x) to get:
r(t) = < -cos(9t)/9 + c₁, -cos(6t)/6 + c₂, 9t²/2 + c₃ >
Where c₁, c₂, and c₃ are the constants of integration.
To find these values, we can use the given restriction:
r(0) = <4, 6, 3> = < -cos(0)/9 + c₁, -cos(0)/6 + c₂, 0 + c₃ >
= < - 1/9 + c₁, -1/6 + c₂, c₃ >
Then we have 3 equations:
first component:
-1/9 + c₁ = 4
then
c₁ = 4 + 1/9 = 37/9
Second component:
- 1/6 + c₂ = 6
then
c₂ = 6 + 1/6 = 37/6
Third component:
c₃ = 3
Now we can conclude that the function is:
r(t) = < -cos(9t)/9 + 37/9, -cos(6t)/6 + 37/6 , 9t²/2 + 3 >
If you want to learn more, you can read:
https://brainly.com/question/24062595
