Ice is forming in a pond at a rate given by dy/dx=ksqrt(t), where y is the thickness of the ice in cm at time t measured in hours since the ice started forming, and k is a positive constant. Find y as a function of t.
dy/dx=ksqrt(t)
dy=ksqrt(t)dt
Integrate both sides:
y=k(2/3)t^(3/2) + C
where C is an integration constant determined from initial conditions.
Thus
y = (2k/3)t^(3/2) + C
To solve for y as a function of t, we need to integrate the given rate equation.
The differential equation is: dy/dx = k√(t)
To solve this, we will separate the variables and integrate both sides.
Begin by rearranging the equation: dy = k√(t)dx
Now, integrate both sides of the equation.
∫ dy = ∫ k√(t)dx
The integral of dy is simply y. For the right side of the equation, we can use the power rule with the substitution u = √(t).
y = ∫ k√(t)dx = k∫ u du
Integrating, we have:
y = k ∫ u du = k * (2/3) u^(3/2) + C
Now, substitute u back in:
y = k * (2/3) (√(t))^(3/2) + C
Simplifying further, we have:
y = k * (2/3) t^(3/4) + C
Therefore, y as a function of t is given by:
y = (2/3)kt^(3/4) + C, where C is the constant of integration.
To find y as a function of t, we need to solve the given differential equation:
dy/dx = k * sqrt(t)
Let's separate the variables and integrate both sides:
1/y dy = k sqrt(t) dx
Integrating both sides:
∫ 1/y dy = ∫ k sqrt(t) dx
ln|y| = 2/3 k (t^(3/2)) + C
where C is the constant of integration.
To solve for y, we exponentiate both sides:
e^(ln|y|) = e^(2/3 k (t^(3/2)) + C)
|y| = e^(2/3 k (t^(3/2))) * e^C
We can absorb the constant into a positive constant:
|y| = Ce^(2/3 k (t^(3/2)))
Since y represents the thickness of the ice, we take the absolute value of y to ensure it remains positive. However, since we're assuming the ice is forming on the top of the pond, we can simplify the equation and drop the absolute value:
y = Ce^(2/3 k (t^(3/2)))
Therefore, the thickness of the ice as a function of time t is given by:
y = Ce^(2/3 k (t^(3/2)))