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Sagot :
To rearrange the formula for the volume of a cone, [tex]\( V = \frac{1}{3} \pi r^2 h \)[/tex], to solve for the radius [tex]\( r \)[/tex], follow these steps:
1. Start with the given formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
2. Isolate [tex]\( r^2 \)[/tex] by first multiplying both sides by 3 to clear the fraction:
[tex]\[ 3V = \pi r^2 h \][/tex]
3. Next, divide both sides by [tex]\( \pi h \)[/tex] to isolate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{3V}{\pi h} \][/tex]
4. Finally, take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi h}} \][/tex]
Since [tex]\( r \)[/tex] represents a radius, which is a length, we typically consider only the positive value. Nevertheless, mathematically, both the positive and negative square roots are solutions.
Thus, the radius [tex]\( r \)[/tex] of the cone is given by:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi h}} \][/tex]
However, when considering the realistic physical context where [tex]\( r \)[/tex] is a length, the solution simplifies to:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
The detailed steps conclude that the two possible values for [tex]\( r \)[/tex] are:
[tex]\[ r = -0.97720502380584 \sqrt{\frac{V}{h}} \quad \text{and} \quad r = 0.97720502380584 \sqrt{\frac{V}{h}} \][/tex]
These two values represent the negative and positive square roots of the term [tex]\(\sqrt{\frac{3V}{\pi h}}\)[/tex] respectively.
1. Start with the given formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
2. Isolate [tex]\( r^2 \)[/tex] by first multiplying both sides by 3 to clear the fraction:
[tex]\[ 3V = \pi r^2 h \][/tex]
3. Next, divide both sides by [tex]\( \pi h \)[/tex] to isolate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{3V}{\pi h} \][/tex]
4. Finally, take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi h}} \][/tex]
Since [tex]\( r \)[/tex] represents a radius, which is a length, we typically consider only the positive value. Nevertheless, mathematically, both the positive and negative square roots are solutions.
Thus, the radius [tex]\( r \)[/tex] of the cone is given by:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi h}} \][/tex]
However, when considering the realistic physical context where [tex]\( r \)[/tex] is a length, the solution simplifies to:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
The detailed steps conclude that the two possible values for [tex]\( r \)[/tex] are:
[tex]\[ r = -0.97720502380584 \sqrt{\frac{V}{h}} \quad \text{and} \quad r = 0.97720502380584 \sqrt{\frac{V}{h}} \][/tex]
These two values represent the negative and positive square roots of the term [tex]\(\sqrt{\frac{3V}{\pi h}}\)[/tex] respectively.
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