Rate of change problem (differentiation)

[Clara Chung]

Homework Statement

Refer to the photo, please verify my answer

Homework Equations

calculus

The Attempt at a Solution

For c, can I do it by assuming Ah=V.
A(dh/dt) + h(dA/dt) = dV/dt then find dA/dt?

 

[Simon Bridge]

Don't you have an expression for A(h)?

 

[Clara Chung]

Don't you have an expression for A(h)?

A(h) =2 (pi) root[1+(h+1)^2] ... then ?

 

[Simon Bridge]

Do you have an expression for V(h) then?

(BTW: very few people will trouble to read images.)

 

[Clara Chung]

Do you have V(h) then?

I don't know what is V(h), I try to assume it as V(h)= A(h)[h] = 2 (pi) root[1+(h+1)^2] h for small interval of h, but I don't know if it is correct.

 

[Simon Bridge]

Didn't you just do a "volume of a solid of revolution" calculation?
Didn't you use the method of disks?

Lessee ... the disk between y and y+dy will have area A(y) and width dy, so it's volume is dV = A(y).dy
You need the volume between the bottom and y=h ... in the problem "h" is not a "small interval", it is a value of y.

For the rest ... you are given the volume flow rate of water dV/dt, and you need dh/dt and dA/dt.
If you know how the height changes with time already, and you know how the area depends on the height ...

 

[Clara Chung]

A(h) =2 (pi) root[1+(h+1)^2], so for c part, can I just differentiate both side, since I know dh/dt = 4/5 in b part.
dA(h)/ dt = 2(pi)(1/2)(1+9)^(-1/2)(2)(2+1) dh/dt
so dA(h)/dt = 24pi/5root(10)

 

[SammyS]

A(h) =2 (pi) root[1+(h+1)^2] ... then ?

That looks like the circumference of the circular surface, not the area.

 

[Clara Chung]

That looks like the circumference of the circular surface, not the area.

A(h) = pi(1+(h+1)^2)
=pi (2+2h+h^2)
dA(h) / dt = pi(2h+2)dh/dt
=24/5
thanks

 

[SammyS]

A(h) = pi(1+(h+1)^2)
=pi (2+2h+h^2)
dA(h) / dt = pi(2h+2)dh/dt
=24/5
thanks

That looks better.See what Simon said:

(BTW: very few people will trouble to read images.)

Notice: It's pretty easy to display your image directly in a post:

 

[Clara Chung]

Thank you for the advice from you both. Can you also teach me how to show that "1-t^2/2 <=cost <=1 for 0<=t<=1 "

 

[SammyS]

Thank you for the advice from you both. Can you also teach me how to show that "1-t^2/2 <=cost <=1 for 0<=t<=1 "

You should start a new thread for this.

 

[Mark44]

@Clara Chung, problems involving differentiation should be posted in the calculus subsection, not the precalculus section.