Ask question

Ask question

All categories

4 years ago

15

If a cylindrical tank holds 100,00 gallons of water which can be drained from thebottom of the tank in an hour, then Torricelli’

s Law gives the volumeVof waterremaining in the tank aftertminutes asV(t) = 100,000(1−160t)20≤t≤60(a) What is the physical meaning ofV′(t)? What are the units?(b) Find and interpret the quantityV′(10).

1 answer:

tatyana61 [14]4 years ago

5 0

Answer:

First, I think the right formula is:

$V(t) = 100,000(1-\frac{t}{60} ),20\leq t\leq 60$

A) We first derive the formula given above:

$V'(t)=-\frac{100000}{60}$

V'(t) represent the drain rate of the tank volume.

B) the units is: gallons/time

C) at V'(10) = 100000/60 = 1666.66 gallons/time.

because the formula V'(t) is constant so it doesn't depend of time.

You might be interested in

8.17 8.073 8.2<br> 8.03 8.063<br> 8.016<br> Least to greatest

Katena32 [7]

Answer:

8.016, 8.03, 8.063, 8.073, 8.17, 8.2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Gus bought a salad for $6 and two pizza's. If he paid a total of $36, what equation represents finding the cost of one pizza?

Jobisdone [24]

Answer:

15 dollars each pizza

Step-by-step explanation:

2 pizzas (15 + 15 = 30)

30 + 6 = 36

4 0

3 years ago

Read 2 more answers

Find the x- and y-intercepts of the graph of each equation. (set x & y equal to 0; two different

bogdanovich [222]

X= 2 y=4 these are the intercepts

5 0

3 years ago

P=a+b+c solve for b and graph the inequality

Nataly [62]

I need points can you give me some of your points

5 0

3 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago