Question

Can anyone help with a calculus problem?

Answer 1

i can

Answer 2

Thank you, I'll post it in a bit.

Answer 3

Suppose a basketball is being inflated with air at the rate of 5cm^3/s find the rate at which the radius of the ball is increasing when the radius is 10 cm.

Answer 4

Has to do with related rates

Answer 5

@jhonyy9 @rebeccaxhawaii @poopsiedoodle

Answer 6

The volume of the basketball is this \[v = \frac{ 4 }{ 3 } \pi*r^{3}*h\]

Answer 7

so i'm assuming that this has to do with the rate the volume is increasing by

Answer 8

yup :))

Answer 9

But you see @rishavraj i'm not sure as to how to put these concepts together

Answer 10

gimme a min :))

Answer 11

\[(\frac{ 4 }{ 3 }*\pi*h)\frac{ dV }{ dr } r^{3} = 3r^{2}*(\frac{ 4 }{ 3 })*\pi*h\]


\[\frac{ dV }{ dr } = 4r^{2}\pi*h\]

\[\frac{ dV }{ ds } = 5cm^{3}/s\]

Answer 12

\[\frac{ dV }{ dt } = 5~and~V = \frac{ 4 }{ 3 }\pi r^3\]

Answer 13

hint:
we can write this:
\[\huge \frac{{dr}}{{dt}} = \frac{{\left( {dV/dt} \right)}}{{4\pi {r^2}}}\]

Answer 14

@greatlife44 can u proceed now??

Answer 15

so, after a substitution, we get:
\[\Large \frac{{dr}}{{dt}} = \frac{{\left( {dV/dt} \right)}}{{4\pi {r^2}}} = \frac{5}{{4\pi \times {{10}^2}}} = ...?\]

Answer 16

question: how were you able to set those two equal to each-other just a little confused

Answer 17

so
\[\frac{ dV }{ dt } = 4 \pi r^2 \frac{ dr }{ dt }\]
u got the vlaues plug them

Answer 18

see
\[V = \frac{ 4 }{ 3 } \pi r^3\]
we jst diffeerentiated it wrt time

Answer 19

more precisely, if we apply the chain rule, we can write this:
\[\huge \frac{{dV}}{{dt}} = \frac{{dV}}{{dr}} \cdot \frac{{dr}}{{dt}} = 4\pi {r^2} \cdot \frac{{dr}}{{dt}}\]

Answer 20

Trying to work through what you said and put this together in my head.

\[\frac{ dV }{ dt } = 5cm^{3}/s\]

\[\frac{ dV }{ dr } = 4*\pi*r^{2}\]

so I get now that we need to find out what the rate the radius is increasing

\[\frac{ dr }{ dt }\]

Answer 21

okay so

\[\frac{ dr }{ dt } = \frac{ \frac{ dv }{ dr } }{ \frac{ dv }{ dt } }\]

Answer 22

yuppp
\[\frac{ dV }{ dt } = 4 \pi r^2 \frac{ dr }{ dt }\]

as dV/dt = 5
so

\[\frac{ dr }{ dt } = \frac{ 5 }{ 4 \pi r^2 }\]
dr/dt is the rate of change of radius

Answer 23

We find how we can express dr/dt in terms of what we already know so is that what you guys are saying.

Answer 24

yes! Please what is:
\[\Large \frac{{dr}}{{dt}} = \frac{{\left( {dV/dt} \right)}}{{4\pi {r^2}}} = \frac{5}{{4\pi \times {{10}^2}}} = ...?\]

Answer 25

\[\frac{ dr }{ dt } = \frac{ \frac{ dv }{ dt } }{ \frac{ dv }{ dr } } = \frac{ 5 }{ 4*\pi*r^{2} }\]

\[\frac{ \frac{ dv }{ dt }*dr~dt }{ \frac{ dv }{ dr }*dr~dt } = \frac{ dr~dv }{ dt~dv } = \]

\[\frac{ 5 }{ 4*\pi*(10)^{2} } = \frac{ 1 }{ 80*\pi } = \frac{ dr }{ dt }\]

Answer 26

so the radius increases at a rate of

\[\frac{ 1 }{ 80*\pi }( cm^{2}/s)\]

Answer 27

that's right!

Answer 28

Thank you again I had to work through what you said before