Conical Cups Related Rates
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My teacher provided the solution and I don't want to just copy it down.
I know we have V=$frac{pi r^2h}{3}$ and $dh/dt = 2$.
What I don't understand is how V=$frac{pi r^2h}{3}$ becomes V=$frac{pi h^3}{3}$. If someone could explain that to me, I will really appreciate it.
Thanks in advance!
calculus geometry
edited Nov 20 at 15:23
Andrei
10.7k21025
asked Nov 20 at 14:58
Lola
156
add a comment |
up vote
0
down vote
favorite
My teacher provided the solution and I don't want to just copy it down.
I know we have V=$frac{pi r^2h}{3}$ and $dh/dt = 2$.
What I don't understand is how V=$frac{pi r^2h}{3}$ becomes V=$frac{pi h^3}{3}$. If someone could explain that to me, I will really appreciate it.
Thanks in advance!
calculus geometry
edited Nov 20 at 15:23
Andrei
10.7k21025
asked Nov 20 at 14:58
Lola
156
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My teacher provided the solution and I don't want to just copy it down.
I know we have V=$frac{pi r^2h}{3}$ and $dh/dt = 2$.
What I don't understand is how V=$frac{pi r^2h}{3}$ becomes V=$frac{pi h^3}{3}$. If someone could explain that to me, I will really appreciate it.
Thanks in advance!
calculus geometry
edited Nov 20 at 15:23
Andrei
10.7k21025
asked Nov 20 at 14:58
Lola
156
My teacher provided the solution and I don't want to just copy it down.
I know we have V=$frac{pi r^2h}{3}$ and $dh/dt = 2$.
What I don't understand is how V=$frac{pi r^2h}{3}$ becomes V=$frac{pi h^3}{3}$. If someone could explain that to me, I will really appreciate it.
Thanks in advance!
calculus geometry
calculus geometry
edited Nov 20 at 15:23
Andrei
10.7k21025
asked Nov 20 at 14:58
Lola
156
edited Nov 20 at 15:23
Andrei
10.7k21025
asked Nov 20 at 14:58
Lola
156
edited Nov 20 at 15:23
Andrei
10.7k21025
edited Nov 20 at 15:23
Andrei
10.7k21025
edited Nov 20 at 15:23
Andrei
10.7k21025
10.7k21025
asked Nov 20 at 14:58
Lola
156
asked Nov 20 at 14:58
Lola
156
asked Nov 20 at 14:58
Lola
156
156
add a comment |
add a comment |
1 Answer
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At the top $r=h=10$ cm. By using similar triangles, you have that $r=h$ at all heights. In general, you would need to write $r$ in terms of $h$, no matter the shape of the cone. In this case is trivial.
answered Nov 20 at 15:21
Andrei
10.7k21025
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
At the top $r=h=10$ cm. By using similar triangles, you have that $r=h$ at all heights. In general, you would need to write $r$ in terms of $h$, no matter the shape of the cone. In this case is trivial.
answered Nov 20 at 15:21
Andrei
10.7k21025
add a comment |
up vote
1
down vote
accepted
At the top $r=h=10$ cm. By using similar triangles, you have that $r=h$ at all heights. In general, you would need to write $r$ in terms of $h$, no matter the shape of the cone. In this case is trivial.
answered Nov 20 at 15:21
Andrei
10.7k21025
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
At the top $r=h=10$ cm. By using similar triangles, you have that $r=h$ at all heights. In general, you would need to write $r$ in terms of $h$, no matter the shape of the cone. In this case is trivial.
answered Nov 20 at 15:21
Andrei
10.7k21025
At the top $r=h=10$ cm. By using similar triangles, you have that $r=h$ at all heights. In general, you would need to write $r$ in terms of $h$, no matter the shape of the cone. In this case is trivial.
answered Nov 20 at 15:21
Andrei
10.7k21025
answered Nov 20 at 15:21
Andrei
10.7k21025
answered Nov 20 at 15:21
Andrei
10.7k21025
answered Nov 20 at 15:21
Andrei
10.7k21025
10.7k21025
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