Question

Calc help on Related Rates and Implicit Differentiation problems Question 1

Answer 1

So what is the formula for the volume of this? Is is \[v=1/3*r^2*h*\pi\]
For part A i did \[1/(b-a) * \int\limits_{a}^{b}r *dr = \]
\[1/(103/20-3/20)*\int\limits_{3/20}^{103/20}* 1/20(3+h^2)dh = .605\]

Answer 2

I got 103/20 from plugging in h =10 into the formula that it gives and got 3/20 from plugging in h=0

Answer 3

@dude @tranquility

Answer 4

Let's try to label that funnel first.

Answer 5

?

Answer 6

@florisalreadytaken

Answer 7

you are looking for b right? what you did above, i dont get its logic -- youve overcomplicated it, and in a wrong way also

Answer 8

so \[r=1/20(3+h^2)\]
so to get the average value for the radius, don't we plug in 0 and 10 in for h (because 0< h< 10 INClusive) into the formula and get "a" and "b" ?

Answer 9

no

b is 10

a is 0

Answer 10

Oh, i see now, let me redo part a

Answer 11

So what is the formula for the volume of this? Is is \[v=1/3*r^2*h*\pi\]
For part A i did \[1/(b-a) * \int\limits_{a}^{b}r *dr = \]
\[1/(10-0)*\int\limits_{0}^{10}* 1/20(3+h^2)dh = 109/60=1.816666\]
*** changed it ****

Answer 12

omg i was doing b and i was like what is this guy doing... and then i looked at the image again and i realised youre doing A -- its 2am so yeah do not expect much

that looks all good

Answer 13

oh alr, so the first was correct or the second for a?

Answer 14

avg value is ok

have you done the rest?

Answer 15

for b use
\[ \pi \int\limits_{a}^{b}r^2 *dr \]

Answer 16

Not quite, so for b to find the volume of the funnel I have to write an eqn for the volume in terms of r and h. But would the eqn be v = 1/3*pi *r^2*h or what
Not sure what the volume formula for a funnel is.......

Answer 17

well yeah you are right its quite a complicated figure

you want to do the same you did with your previous post right? separate it into circles

do you get he idea of it?

Answer 18

treat it like a cylinder

Answer 19

One last thing i want to point out be4 i go haha

I meant dh not dr

That would be the height

I just copied your latex thats why its dr lol : )

Answer 20

How to do part c though?
@az maybe if you have time

Answer 21

The volume of the funnel is going to be the sum of the areas of each consecutive circle that's formed as a cross section at different heights

so
\[ \text{Volume} = \pi \int\limits_{a}^{b}r^2 ~dh \]

Volume = pi * r^2 h

\(\pi\) is a constant so it's taken out and then you're integrating it with respect to the height so dh

Answer 22

For the related rates part c,

we know
r = 1/20 (3 + h^2)

you can distribute the 1/20 inside and we get

\( r = \dfrac{3}{20} + \dfrac{h^2}{20}\)

now it's easier to differentiate and you won't make any silly mistakes

We want to know the rate at which the height is decreasing with respect to time (that is dh/dt)

we know r = whatever

can you differentiate that with respect to time

so what is dr/dt = ??

and then you have to use what the question states, they tell us that the radius of the liquid is decreasing at a rate of 1/5 inch per second
so that means you substitute -1/5 in place of dr/dt

and then all you have to do is solve for dh/dt and that's your final answer

Answer 23

\[r=3/20+h^2/20\]
\[dr/dt=0+h/10*dh/dt\]
-1.5 = 0 + 3/10 * dh/dt
-1.5*10/3 = dh/dt

Answer 24

\(\dfrac{\text{dr}}{\text{dt}} = \dfrac{d}{\text{dt}} \left(\dfrac{3}{20} + \dfrac{h^2}{20}\right) = 0 + \left( \dfrac{1}{20} \cdot\dfrac{d}{\text{dt}} \left(h^2\right) \right) \)

\( = \dfrac{1}{20} \cdot 2h\cdot \dfrac{\text{dh}}{\text{dt}} \)

\(\dfrac{\text{dr}}{\text{dt}} = \dfrac{h}{10} \cdot \dfrac{\text{dh}}{\text{dt}}\)

so, yup good job

but be careful, it's -1/5
not -1.5

but other than that, you're on the right track now :)

Answer 25

oh alr, yeah My head, I "mis-saw" that lol. Thanks (:

Answer 26

You're welcome :)