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: 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).

Answer 2

`\boxed{\boxed{y(t)=(2√(k))/(3)t^{(3)/(2)}+c}}`

Explanation:

Rate of ice forming on the pond is,

`(dy)/(dt)=√(kt)`

where,

y = the thickness of the ice in centimeters

t = time passed in hours

k = positive constant

Then,

`dy=√(kt)\cdot dt`

Taking integrals on both sides,

`\int dy=\int √(kt)\cdot dt`

`\int dy=\int √(k)\cdot √(t)\cdot dt`

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

`\int dy=√(k)\int √(t)\cdot dt`

`y=√(k) \frac{t^{(3)/(2)}}{(3)/(2)}+c`

`y=(2√(k))/(3)t^{(3)/(2)}+c`