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Studying for a test? Prepare with these 2 lessons on Module 5: Examples of functions from geometry.
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Video-Transkript
Frank wants to fill up a spherical water balloon with as much water as possible. The balloons he bought can stretch to a radius of 3 inches-- not too big. If the volume of a sphere is-- and this is volume as a function of radius-- is equal to 4/3 pi r cubed, what volume of water in cubic inches can Frank put into the balloon? So this function definition is going-- if you give it a radius in inches, it's going to produce a volume in cubic inches. So let's rewrite it. Volume as a function of radius is equal to 4/3 pi r cubed. Now, they say the balloons he bought can stretch to a radius of 3 inches. So let's think about, if the radius gets to 3 inches, what the volume of that balloon is going to be. So we essentially would just input 3 inches into our function definition. So everywhere where we see an r, we would replace it with a 3. So we could write-- and just to be clear, let me rewrite it in the same color. V of-- that's not the same color. We do it in that brownish color right over here. So V of 3 is equal to 4/3 pi-- and instead of r cubed, I would write 3 cubed-- 4/3 pi 3 cubed. This is how the function definition works. Whatever we input here, it will replace the r in the expression. So V of 3 is going to be equal to-- so this is going to be equal to 4/3 pi times-- 3 to the third power is 27. 27 divided by 3 is 9, so this is 9. 9 times 4 is 36 pi. So this is equal to 36 pi. And since this was in inches, our volume is going to be in inches cubed or cubic inches. So that's the volume of water that Frank can put in the balloon-- 36 pi cubic inches.