Mathematics: The volume v of a right circular cylinder of radius r and heigh h is v = pi r^2 h 1. how is dv/dt related to dr/dt if h is constant and r varies with time? 2. how is dv/dt related to dh/dt if r is constant and h varies with time? 3. how is dv/dt related to dh/dt and dr/dt if both h and r vary with time?

The volume v of a right circular cylinder of radius

Ответ:

multibeastpr

13.03.2022

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case (

and

not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

0,0(0 оценок)

Ответ:

kemellgwilson

08.06.2022

Mathematics

p = 3

q = 2

Step-by-step explanation:

they have infinitely many solutions, if they are basically the same equations. in other words, when they can be combined/reduced into a statement of tautology, a statement that is always true, no matter what.

like 15 = 15 or 8x = 8x

now we need to aim for creating the same (or multiples of the same) factors in the original equation 2 as in the original equation 1.

in other words we find the p and q, where we get the same ratios between the factors of squadron 1 and equation 2.

2/(p+q+1) = 3/(p+2q+2) = 7/(4p +4q +1)

p+q+1 = 2/3 × (p+2q+2) = 2/7 × (4p + 4q + 1)

3p + 3q + 3 = 2p + 4q + 4 = 6/7 × (4p+4q+1)

21p + 21q + 21 = 14p + 28q + 28 = 24p + 24q + 6

7p - 7q = 7

p - q = 1

p = 1 + q

14(1+q) + 28q + 28 = 24(1+q) + 24q + 6

14 + 14q + 28q + 28 = 24 + 24q + 24q + 6

42 + 42q = 30 + 48q

12 = 6q

q = 2

p = 1 + 2 = 3

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question