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, $\frac{dy}{dt}$, 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.
${\color{blue}\frac{dy}{dt} \:=\:k\sqrt{t}}$

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

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

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

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

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

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

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

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

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). :-)