# Thread: Calc 3 question: initial value problem

1. ## Calc 3 question: initial value problem

Hello,

I'm not sure how to get started on this:

Solve initial value problem for r as a function of t

Differential equation: d^2r/dt^2 = -(i + j + k)
Initial conditions: r(0) = 10i +10j +10k and (dr/dt) evaluated at t=0 =0

I'd appreciate all the help I can get; truthfully, I'm pretty lost in this course and regretting taking it at all.

2. Originally Posted by riverjib
Hello,

I'm not sure how to get started on this:

Solve initial value problem for r as a function of t

Differential equation: d^2r/dt^2 = -(i + j + k)
Initial conditions: r(0) = 10i +10j +10k and (dr/dt) evaluated at t=0 =0

I'd appreciate all the help I can get; truthfully, I'm pretty lost in this course and regretting taking it at all.
what are the initial conditions for r'(0) ?

3. Hello, riverjib!

Solve initial value problem for $r$ as a function of $t.$

Differential equation: . $\frac{d^2r}{dt^2} \:=\:-(i + j + k)$
Initial conditions: . $r(0) \:=\:10i +10j +10k,\quad \frac{dr}{dt}(0) \:=\:0$

We have: . $\frac{d^2r}{dt^2} \;=\;-(i+j+k)$

Integrate: . $\frac{dr}{dt} \;=\;-(i+j+k)t + C_1$

We are told that, when $t = 0,\:\frac{dr}{dt} = 0$

. . So we have: . $-(i+j+k)0 +C_1 \:=\:0\quad\Rightarrow\quad C_1 = 0$

Hence: . $\frac{dr}{dt} \:=\:-(i+j+k)t$

Integrate: . $r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + C_2$

We are told that, when $t = 0,\:r = 10i + 10j +10k$

. . So we have: . $-\frac{1}{2}(i+j+k)0^2 + C_2 \:=\:10i + 10j + 10k\quad\Rightarrow\quad C_2\:=\:10i + 10j + 10k$

Therefore: . $\boxed{r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + 10(i+j+k)}$

4. Originally Posted by Soroban
Hello, riverjib!

We have: . $\frac{d^2r}{dt^2} \;=\;-(i+j+k)$

Integrate: . $\frac{dr}{dt} \;=\;-(i+j+k)t + C_1$

We are told that, when $t = 0,\:\frac{dr}{dt} = 0$

. . So we have: . $-(i+j+k)0 +C_1 \:=\:0\quad\Rightarrow\quad C_1 = 0$

Hence: . $\frac{dr}{dt} \:=\:-(i+j+k)t$

Integrate: . $r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + C_2$

We are told that, when $t = 0,\:r = 10i + 10j +10k$

. . So we have: . $-\frac{1}{2}(i+j+k)0^2 + C_2 \:=\:10i + 10j + 10k\quad\Rightarrow\quad C_2\:=\:10i + 10j + 10k$

Therefore: . $\boxed{r(t) \;=\;-\frac{1}{2}(i+j+k)t^2 + 10(i+j+k)}$

oh! i didn't notice the part where it said dr/dt = 0

,

,

### solve initial value problem dr/dt tr = 0 ; condition r(0) = r0

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