1. differencial equation

The rate, dy/dt, at which ice is forming on a pond is proportional to the square root of t. Let y be the thickness of the ice in inches at time t measured in hours since the ice started forming.

(a) Write the differential equation. Use k > 0 as the proportionality constant.
dy/dt =

(b) Solve the differential equation.
y(t) =

(c) If the thickness of the ice is 3 inches after 9 hours, what is the value of k?
k =

a little help with these bad frickin johnnys...thanks

mathlete

2. Hello, mathlete!

The rate, $\displaystyle \frac{dy}{dt}$, at which ice is forming on a pond
is proportional to the square root of $\displaystyle t.$
Let $\displaystyle y$ be the thickness of the ice in inches at time $\displaystyle t$
measured in hours since the ice started forming.

(a) Write the differential equation. Use $\displaystyle k > 0$ as the proportionality constant.
$\displaystyle {\color{blue}\frac{dy}{dt} \:=\:k\sqrt{t}}$

(b) Solve the differential equation.
We have: .$\displaystyle \frac{dy}{dt}\;=\;k\!\cdot\!t^{\frac{1}{2}}$

Separate variables: .$\displaystyle dy \;=\;k\!\cdot\!t^{\frac{1}{2}}dt$

Integrate: .$\displaystyle y \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}} + C$

Assuming there was no ice at $\displaystyle t = 0$, we have: .$\displaystyle C \:=\:0$

The ice-thickness function is: .$\displaystyle {\color{blue}y \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}}}$

(c) If the thickness of the ice is 3 inches after 9 hours,
what is the value of $\displaystyle k$?
We are told: .when $\displaystyle t = 9,\:y = 3$

So we have: .$\displaystyle 3 \;=\;\frac{2}{3}k\!\cdot\!9^{\frac{3}{2}} \;=\;\frac{2}{3}k\cdot27\quad\Rightarrow\quad18k \:=\:3$

. . Therefore: .$\displaystyle {\color{blue}k \:=\:\frac{1}{6}}$

3. The rate, dy/dt, at which ice is forming on a pond is proportional to the square root of t. Let y be the thickness of the ice in inches at time t measured in hours since the ice started forming.

(a) Write the differential equation. Use k > 0 as the proportionality constant.
dy/dt = k*sqrt(t)

(b) Solve the differential equation.
y(t) =
dy = k(t)^(1/2) dt
y = (1/2)k[t^(-1/2) +C
y = (k/4)sqrt(t) +C

When t=0, y=0, so,
0 = (k/4)sqrt(0) +C
C = 0

Therefore, y(t) = (k/4)sqrt(t)
(c) If the thickness of the ice is 3 inches after 9 hours, what is the value of k?
k =
3 = (k/4)sqrt(9)
3 = (k/4)*3
k = 4

EDIT: Uh-oh! After seeing Soroban's solution, I admit mine is wrong. Wrong Integration of sqrt(t). :-)