Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin9t, sin6t,9t⟩,r(0)=⟨4,6,3⟩

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:

and:

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:

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With the initial condition , we find

So the particular solution to the IVP is

(c, 0) i might be wrong so don't put full faith on my answer

2.

according to newton’s cannonball model, what happens if the cannonball's velocity is enough to overpower the pull of earth’s gravitystep-by-step explanation: