Finding volume?

[Pimptatay]

I got another problem. And I wanna make sure I’m doing this right.
There’s no context again, just write formula and solve.

Problem: “The ratio of volume of two spheres is equal to the ratio of the cubes of their diameters. The volume of one sphere is 24.25 cm³. Find the volume of another sphere whose diameter is twice as much.”

here’s what I got so far...

 

[Romsek]

you made it so hard....

The ratio of volumes is equal to the ratio of the cubes of their diameters.
If the ratio of diameters is 2, then the ratio of their cubes is \(\displaystyle 2^3=8\)

So just multiply the first volume by 8. No need for the rest of it.

 

[lev888]

The answer is correct, but the solution is unnecessarily complicated and not what the question prompts you to do.

 

[Pimptatay]

I guess I over think things. Thank you ?

 

[Steven G]

I got another problem. And I wanna make sure I’m doing this right.
There’s no context again, just write formula and solve.

Problem: “The ratio of volume of two spheres is equal to the ratio of the cubes of their diameters. The volume of one sphere is 24.25 cm³. Find the volume of another sphere whose diameter is twice as much.”

here’s what I got so far...

View attachment 14697

I disagree with the other posters. You did not get the correct answer. The answer, as already pointed out is, 24.25*8cm^3 which is NOT 194.13cm^3. That answer may be close, but it is not correct.
Also your 2nd line is not correct. You can't divide the left side by 4/(3pi) and do the same to the right side!

 

[Dr.Peterson]

There’s no context again, just write formula and solve.

Problem: “The ratio of volume of two spheres is equal to the ratio of the cubes of their diameters. The volume of one sphere is 24.25 cm³. Find the volume of another sphere whose diameter is twice as much.”

You don't need to know the formula for volume of a sphere to solve this; they told you everything you need (and you should know what they told you anyway). It is true for ANY similar figures, that the ratio of volumes is the ratio of the cubes of any linear dimension.

In the specific case of spheres, this is because [MATH]\frac{V_2}{V_1} = \frac{\frac{4}{3} \pi r_2^3}{\frac{4}{3} \pi r_1^3} = \frac{r_2^3}{r_1^3} = \frac{(D_2/2)^3}{(D_1/2)^3} = \frac{D_2^3}{D_1^3}[/MATH]. But that is true of any volume formula.

To do what they told you to do, just use the fact they stated, [MATH]V_2 : V_1 = D_2^3 : D_1^3[/MATH]. Since you know that [MATH]V_1 = 24.25[/MATH] and [MATH]D_1 : D_2 = 2[/MATH]. Then [MATH]V_2 : V_1 = D_2^3 : D_1^3= (D_2 : D_1)^3 = 2^3 = 8[/MATH], so [MATH]V_2 = 8V_1 = 24.25*8 = 194[/MATH] (exactly -- your rounding caused your answer to be off).

By the way, there is always a context! For example, is this from a course you are taking? What topics are you learning?

 

[Pimptatay]

You don't need to know the formula for volume of a sphere to solve this; they told you everything you need (and you should know what they told you anyway). It is true for ANY similar figures, that the ratio of volumes is the ratio of the cubes of any linear dimension.

In the specific case of spheres, this is because [MATH]\frac{V_2}{V_1} = \frac{\frac{4}{3} \pi r_2^3}{\frac{4}{3} \pi r_1^3} = \frac{r_2^3}{r_1^3} = \frac{(D_2/2)^3}{(D_1/2)^3} = \frac{D_2^3}{D_1^3}[/MATH]. But that is true of any volume formula.

To do what they told you to do, just use the fact they stated, [MATH]V_2 : V_1 = D_2^3 : D_1^3[/MATH]. Since you know that [MATH]V_1 = 24.25[/MATH] and [MATH]D_1 : D_2 = 2[/MATH]. Then [MATH]V_2 : V_1 = D_2^3 : D_1^3= (D_2 : D_1)^3 = 2^3 = 8[/MATH], so [MATH]V_2 = 8V_1 = 24.25*8 = 194[/MATH] (exactly -- your rounding caused your answer to be off).

By the way, there is always a context! For example, is this from a course you are taking? What topics are you learning?

In that case yes, it is from a course I’m taking. Math with application to science and technology. It’s building me up for my next course called Ship’s Stability. And then physics next year, again for higher Ship’s stability and possibly other navigation purposes.