The radius, r, in centimeters, of a melting snowball is given by r=50-2.5t, where t is time in hours. the snowball is spherical, with volume V=4/3(pi)r^3 cm^3. Find a formula for V=f(t), the volume of the snowball as a function of time.
Well, if we're dealing with a snowball, we might as well have some fun with it! So, here's the formula for the volume of the snowball as a function of time:
V(t) = (4/3) * pi * (50 - 2.5t)^3 cm³
Now that's a snowball with potential!
To find a formula for V as a function of time, we substitute the expression for r into the volume formula.
Given that r = 50 - 2.5t, we can substitute this into the volume formula V = (4/3)(π)(r^3) to get:
V = (4/3)(π)(50 - 2.5t)^3
Simplifying further:
V = (4/3)(π)(125000 - 3750t + 37.5t^2 - 0.125t^3)
Thus, the formula for V as a function of time is:
V = 166666.67π - 12500πt + 125πt^2 - (4/3)πt^3
To find a formula for the volume of the snowball as a function of time, we need to substitute the expression for the radius into the formula for the volume.
Given that the radius, r, is given by r = 50 - 2.5t, we can substitute this expression into the formula for the volume:
V = (4/3) * π * r^3
Replacing r with the expression 50 - 2.5t:
V = (4/3) * π * (50 - 2.5t)^3
Now, we can simplify the expression to get the formula for V as a function of time. Let's start by expanding the cube:
V = (4/3) * π * (50 - 2.5t) * (50 - 2.5t) * (50 - 2.5t)
V = (4/3) * π * (50 - 2.5t) * (2500 - 250t + 6.25t^2)
Next, we can distribute and simplify further:
V = (4/3) * π * (125000 - 12500t + 312.5t^2 - 125t + 12.5t^2 - 0.3125t^3)
V = (4/3) * π * (125000 - 13750t + 324.5t^2 - 0.3125t^3)
Finally, we can arrange the terms in descending order of powers of t and express it as a function of time:
V = (4/3) * π * (-0.3125t^3 + 324.5t^2 - 13750t + 125000)
Therefore, the formula for V as a function of time is:
V(t) = (4/3) * π * (-0.3125t^3 + 324.5t^2 - 13750t + 125000)
Just sub the value of r into the volume formula
V = (4/3)π(50 - 2.5t)^3
expand if necessary, I wouldn't.