Answer:
A. 90.5 cubic meters pet second.
B. 3.77 square meters per second.
C. [tex]\frac{dV}{dS} = \frac{r}{2}[/tex]
Step-by-step explanation:
The radius of a sphere is increasing at a constant rate of 0.05 meters per second.
Therefore, [tex]\frac{dr}{dt} = 0.05[/tex] ......... (1)
A. Now, volume of the sphere is given by
[tex]V = \frac{4}{3}\pi r^{3}[/tex]
Now, differentiating both sides with respect to t we get,
[tex]\frac{dV}{dt} = \frac{4}{3}\pi (3r^{2}) \frac{dr}{dt} = 4\pi r^{2} \frac{dr}{dt}[/tex]
Then at r = 12 meters, the rate of increase in volume will be
[tex]\frac{dV}{dt} = 4 \times (\frac{22}{7}) \times 12^{2} \times 0.05 = 90.5[/tex] cubic meters pet second. {From equation (1)}
B. When the volume of the sphere is 36π cubic meters, then
[tex]\frac{4}{3}\pi r^{3} = 36\pi[/tex]
⇒ [tex]r^{3} = 27[/tex]
⇒ r = 3 meters.
Now, surface area of a sphere is given by
S = 4πr² .......... (2)
Differentiating both sides with respect to time (t) we get,
[tex]\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8 \times (\frac{22}{7}) \times 3 \times 0.05 = 3.77[/tex] square meters per second. {From equation (1)}
C. Now, [tex]V = \frac{4}{3}\pi r^{3}[/tex] and S = 4πr²
⇒ [tex]V = \frac{1}{3} (4\pi r^{2})r = \frac{Sr}{3}[/tex]
Now, differentiating with respect to S both sides we get,
[tex]\frac{dV}{dS} = \frac{r}{3} + \frac{S}{3} \frac{dr}{dS}[/tex] ......... (3)
Now, we have, S = 4πr²
Differentiating with respect to S both sides, we get
[tex]1 = 8\pi r\frac{dr}{dS}[/tex]
⇒ [tex]\frac{dr}{dS} = \frac{1}{8\pi r }[/tex] ......... (4)
Now, from equations (2), (3) and (4) we get,
[tex]\frac{dV}{dS} = \frac{r}{3} + \frac{4\pi r^{2} }{3} \times \frac{1}{8\pi r }[/tex]
⇒ [tex]\frac{dV}{dS} = \frac{r}{3} + \frac{r}{6}[/tex]
⇒ [tex]\frac{dV}{dS} = \frac{r}{2}[/tex] (Answer)
Question:
C. Express the rate at which the volume of the sphere changes with respect to the surface are of the sphere (as a function of r)
Step-by-step explanation:
Here's how you solve part C of this question.
Start with what you know:
dV/dt = 4pir^2 x (dr/dt)
(dr/dt) = 0.05
S(r) = (Surface Area) = 4pir^2
So, since S(r) is equal to 4pir^2, we replace 4pir^2 with S(r). We also know that (dr/dt) = 0.05, so we replace (dr/dt) with 0.05.
Now the equation should look like this:
dV/dt = 0.05 x S(r) <----- (your answer).
Hope this helped!! :)
