Question

Ice is forming on a pond at a rate given by dydt=kt√, dydt=kt, where yy is the thickness of the ice in centimeters at time tt measured in hours since the ice started forming, and kk is a positive constant. find yy as a function of tt.

Answer 1

Answer:

$\boxed{\boxed{y(t)=\frac{2\sqrt{k}}{3}t^{\frac{3}{2}}+c}}$

Step-by-step explanation:

Rate of ice forming on the pond is,

$\dfrac{dy}{dt}=\sqrt{kt}$

where,

y = the thickness of the ice in centimeters

t = time passed in hours

k = positive constant

Then,

$dy=\sqrt{kt}\cdot dt$

Taking integrals on both sides,

$\int dy=\int \sqrt{kt}\cdot dt$

$\int dy=\int \sqrt{k}\cdot \sqrt{t}\cdot dt$

As k is a constant, it can come outside the integral,

$\int dy=\sqrt{k}\int \sqrt{t}\cdot dt$

$y=\sqrt{k} \dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}+c$

$y=\frac{2\sqrt{k}}{3}t^{\frac{3}{2}}+c$

Answer 2

Answer: Multiply both sides by dt, yielding... dy = kt^(1/2)*dt. Integrate both sides, left with respect to y and right with respect to t, using the power rule. y = k(2/3)t^(3/2). y = (2k/3)*t^(3/2).