Step-by-step explanation:
we need to use both equations for the volumes.
the only variables in them are r1 and r2.
when we say both results (volumes) are equal, that means we can create an equation with one formula on one side and the other formula on the other side.
and then we can easily transform the equation into "r1 = ..." or "r2 = ..."
and yes, your teacher made a mistake by hinting to solve for r1 in a. and b. it is r1 in a. and r2 in b.
a.
the goal here is an equation in the form
"r1 = ..."
that means r1 is a function of r2.
36×pi×r1² = 1/2 × 4/3 × pi × r2³ = 4/6 × pi × r2³ =
= 2/3 × pi × r2³
36 × r1² = 2/3 × r2³
r1² = (2/3 / 36) × r2³ = (1/(3×18)) × r2³ = 1/54 × r2³
r1 = sqrt(1/54 × r2³) = 1/3 × sqrt(1/6 × r2³)
b.
the goal here is
"r2 = ..."
again
36×pi×r1² = 2/3 × pi × r2³
36 × r1² = 2/3 × r2³
3×36 × r1² = 2 × r2³
3×18 × r1² = r2³
54 × r1² = r2³
![r2 = \sqrt[3]{54 * {r1}^(2) } = 3 * \sqrt[3]{2 * {r1}^(2) }](https://img.qammunity.org/2024/formulas/mathematics/high-school/2btqrtsbgtrrd8vyjamlh6xzyy68rwue4s.png)
c.
r2 is doubled.
so, we have to go back to our equation of a.
r1 = sqrt(1/54 × r2³).
when r2 is doubled, we get
r1 = sqrt(1/54 × (2×r2)³) = sqrt(1/54 × 8 × r2³) =
= sqrt(8) × sqrt(1/54 × r2³)
so, when r2 doubles, r1 has to grow by the factor of sqrt(8) to keep the equality intact.