Solve the initial value problem for r as a vector function of t Differential equation: drdt=(180t)i+(180t−16t2)j Initial condition: r(0)=100j

Step 1 Given differential equation, drdt=(180t)i+(180t−16t2)j Initial condition, r(0)=100j Step 2 Differential equation can be written as dr=[(180t)i+(180t−16t2)j]d On integrating both sides, r(t)=∫[(180t)i+(180t−16t2)j]dt=180×t22i+[180×t22i−16×t33i]j+C here is an integration constant. r(t)=90t2i+[90t2−163t3]j+C Step 3 Apply the initial condition, r(0)=100j r(0)=90t2i+[90t2−163t3]j+C 100j=90(0)i+[90(0)−163(0)]j+C C=100j Therefore, r(t)=90t2i+[90t2−163t2]j+100j r(t)=90t2i+[90t2−163t2+100]j Step 4 Hence, solution of initial value problem is r(t)=90t2i+[90t2−163t2+100]j

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High School
Algebra
Algebra I
Quadratic function and equation

rabbitz42z8

Answered question

2021-11-30

Solve the initial value problem for r as a vector function of t Differential equation:

\frac{d r}{d t} = (180 t) i + (180 t - 16 t^{2}) j

Initial condition:

r (0) = 100 j

Answer & Explanation

Ralph Lester

Beginner2021-12-01Added 16 answers

Step 1
Given differential equation,
$\frac{d r}{d t} = (180 t) i + (180 t - 16 t^{2}) j$
Initial condition,
$r (0) = 100 j$
Step 2
Differential equation can be written as
$d r = [(180 t) i + (180 t - 16 t^{2}) j] d$
On integrating both sides,
$r (t) = \int [(180 t) i + (180 t - 16 t^{2}) j] d t = 180 \times \frac{t^{2}}{2} i + [180 \times \frac{t^{2}}{2} i - 16 \times \frac{t^{3}}{3} i] j + C$
here is an integration constant.
$r (t) = 90 t^{2} i + [90 t^{2} - \frac{16}{3} t^{3}] j + C$
Step 3
Apply the initial condition,
$r (0) = 100 j$
$r (0) = 90 t^{2} i + [90 t^{2} - \frac{16}{3} t^{3}] j + C$
$100 j = 90 (0) i + [90 (0) - \frac{16}{3} (0)] j + C$
$C = 100 j$
Therefore,
$r (t) = 90 t^{2} i + [90 t^{2} - \frac{16}{3} t^{2}] j + 100 j$
$r (t) = 90 t^{2} i + [90 t^{2} - \frac{16}{3} t^{2} + 100] j$
Step 4
Hence, solution of initial value problem is
$r (t) = 90 t^{2} i + [90 t^{2} - \frac{16}{3} t^{2} + 100] j$

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