can i derive this equation. s(v)=(4pi)^1/3(3v)^2/3 and please explain.
To derive the equation given, s(v) = (4π)^(1/3)(3v)^(2/3), we can go through the process of simplifying and applying some rules of calculus. Here's how you can do it step by step:
Step 1: Expand the equation
Start by expanding the equation using the properties of exponents. Rewrite (4π)^(1/3) as the cube root of (4π) and (3v)^(2/3) as the cube root of (3v)^2.
s(v) = (4π)^(1/3)(3v)^(2/3)
= (4π)^(1/3) * (3v)^(2/3)
= (4π)^(1/3) * [(3v)^2]^(1/3)
Step 2: Apply the power rule of exponents
The power rule states that when you raise a power to an exponent, you multiply the exponents. Applying this rule will simplify the equation further.
s(v) = (4π)^(1/3) * [(3v)^2]^(1/3)
= (4π)^(1/3) * (3^(2/3) * v^(2/3))
Step 3: Simplify the expression
To simplify the expression, combine the terms and simplify any constants.
s(v) = (4π)^(1/3) * (3^(2/3) * v^(2/3))
= (4π)^(1/3) * 3^(2/3) * v^(2/3)
= (4π * 3^(2/3))^(1/3) * v^(2/3)
This is the simplified form of the equation s(v) = (4π)^(1/3)(3v)^(2/3).
So, by expanding the original equation and applying the power rules of exponents, we simplified it to (4π * 3^(2/3))^(1/3) * v^(2/3).