## marigirl one year ago Rate of change -Differentiation An inverted cone with a base area of 2500pi mm^2 and height of 300m is slowly being filled with water. Water is being poured in at a rate of 350 mm^3/s. At what rate is the height of the water changing when the water is 100mm from the top of the inverted cone?

1. marigirl

I think: We are to find out: $\frac{ dh }{ dt }$ We know $\frac{ dV }{ dt }=350$

2. marigirl

|dw:1443476761216:dw|

3. marigirl

Volume of cone: $V=\frac{ 1 }{ 3 }\pi r^2h$ Volume of this cone:$V=\frac{ 2500 }{ 3 }h$

4. phi

use "similar triangles" to write r in terms of h

5. phi

from the area of the base you find r=50 and you have this |dw:1443477912255:dw|

6. phi

$V=\frac{ 1 }{ 3 }\pi r^2h \\ V=\frac{ 1 }{ 3 }\pi \frac{h^2}{36}h \\ V= \frac{\pi}{108} h^3$ now use implicit differentiation

7. marigirl

$\frac{ dv }{ dh }=36 \pi h^2$ $\frac{ dh }{ dt }=\frac{ dv }{ dt }\times \frac{ dh }{ dv }$ am i on the right track

8. marigirl

then $\frac{ dh }{ dt }=350 \times \times \frac{ 1 }{ 36 \pi h^2 }$ and then sub in h=100

9. phi

I would take the derivative of each variable with respect to t on the left side you get dv/dt on the right side you get $\frac{\pi}{108} \frac{d}{dt} h^3$ which is $\frac{\pi}{108} 3 h^2\frac{d}{dt}h$ so the equation is $\frac{dV}{dt}= \frac{\pi}{36} h^2 \frac{dh}{dt}$

10. phi

when the water is 100mm from the top of the inverted cone in the formula h is the height from 0 up to 300 100 mm from the top corresponds to h= 200 mm

11. phi

Your approach works except that this ***$$\frac{ dv }{ dh }=36 \pi h^2$$*** should read $\frac{dV}{dh}= \frac{\pi}{36} h^2$

12. marigirl

then sub in h=200....

13. phi

yes

14. marigirl

Thanks heaps!