Related Rates word problems x3

[Guest]

Hi I have a few questions I don't seem to be understanding, help please? =)

What is the first time after 3 o'clock that the hands (hour and minute) of the clock are together?

wow I dont get how to find the varibles for this and what is what..

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The volume of a right circular cylinder is 60cm^3 and is increasing at 2 cm^3/min at a time when the radius is 5cm and is increasing at 1cm/min. How fast is the height of the cylinder changing at that time?

I did this: v=60cm^3; dv/dt= 2cm^3/min;r=5cm,dr/dt=1cm/min
dh/dt=?

v=Pir^2h
60=pi(5^2)h
60/25pi=h
12/5pi=h

v=Pir^2h <--taking derivative with repect to time
dv/dt= pi2r(dr/dt)h+r^2(dh/dt)
2=pi(2)(5)(1)(12/5pi)+(5)^2(dh/dt) <--so I canceled the 5pis out, can't u do that?
2=24+25(dh/dt)
-22/25=dh/dt

therefore ans.= -22/25 cm/min. But the ans is suppose to be -22/25pi cm.min, so who's right?


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A water tank is in the shape of an inverted right circular cone with top radius 10 m and depth 8 m. Water is flowing in at a rate of 1/10m^3/min. How fast is the depth of water in the tank increasing when the water is 4m deep?

Im confused on what is what, cuz if depth is height, why do they give u 2 different heights.
I put this down r=10; h= 8m; dv/dt= 1/20m^3; h=4m
dh/dt=?

v=Pir^2h/3 and what is a Inverted right circular cone? do u still use the volume of a cone formula?

thanks for the help!

 

[galactus]

For the cone:

Find \(\displaystyle \L\\\frac{dh}{dt}\) given that \(\displaystyle \frac{dV}{dt}=20\)

You know the volume of a cone formula.

Use similar triangles:

\(\displaystyle \L\\\frac{r}{h}=\frac{10}{8}\)

Solve for r:

\(\displaystyle \L\\r=\frac{5h}{4}\)

Now, sub that into your cone formula, Differentiate with respect to time(dh/dt). You know dV/dt=20, so solve for dh/dt, sub in your given value for h and voila!.



For the cylinder:

Differentiate the volume of cylinder formula.

\(\displaystyle \L\\\frac{dV}{dt}={\pi}(r^{2}\frac{dh}{dt}+2rh\frac{dr}{dt})\)

Using the formula for the volume of a cylinder, solve for h and sub into dV/dt.

Now, sub in your values and solve for dh/dt.

 

[tkhunny]

What is the first time after 3 o'clock that the hands (hour and minute) of the clock are together?

1) Think about the Domain. The hour hand will be between 3 and 4, right? This substantially narrows the possible results that you have to ponder.

2) Think speed.
Minute Hand Angular Velocity = 2pi radians/hour
Hour Hand Angular Velocity = pi/12 radians/hour

3) Think Starting Position.
Minute Hand Starts at pi/2
Hour Hand Starts at 0

4) Model the Locations as a function of Time, t hours
Minute Hand: M(t) = pi/2 - (2pi)*t
Hour Hand: H(t) = 0 - (pi/12)*t

5) There had better be a solution 0 > H(t₀)=M(t₀) > -pi/12
pi/2 - (2pi)*t = 0 - (pi/12)*t
pi/2 = (2pi - pi/12)*t
t = (pi/2)/(2pi - pi/12) = (1/2)/(2 - 1/12) = (1/2)/(23/12) = 6/23 hours

6) Check
Minute Hand: M(t) = pi/2 - (2pi)*(6/23) = pi/2 - (12/23)pi = -pi/46
Hour Hand: H(t) = 0 - (pi/12)*(6/23) = -pi/46
0 > -pi/46 > -pi/12 <== Good!

7) Understand: 6/23 hours = 15.65217391 minutes = 15 min 39.130435 sec

8) Wonder: Why is this a calculus problem?

 

[soroban]

Hello, bittersweet!

There are several ways to solve the first one.
\(\displaystyle \;\;\)Here's the one I prefer . . .

What is the first time after 3 o'clock that the hands (hour and minute) of the clock are together?

The minute hand moves 360° in one hour . . . or 6° per minute.

The hour hand moves \(\displaystyle 30^o\) an hour . . . or \(\displaystyle \frac{1}{2}^o\) per minute.

The minute hand starts at \(\displaystyle 0^o\).
\(\displaystyle \;\;t\) minutes later, it is at: \(\displaystyle \,0\,+\,6t\) degrees.

The hour hand starts at \(\displaystyle 90^o\).
\(\displaystyle \;\;t\) minutes later, it is at: \(\displaystyle \,90\,+\,\frac{1}{2}t\) degrees.

Since the hand are together: \(\displaystyle \,6t\;=\;90\,+\,\frac{1}{2}t\;\;\) . . . There!


Solve for \(\displaystyle t\) and get: \(\displaystyle \,t\:=\:\frac{180}{11}\:=\:16\frac{4}{11}\) minutes.