Lesson Finding rate of change of some processes

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Finding rate of change of some processes


Problem 1

A spherical snow ball is melting at  120  cm^3/min.
At what rate is the surface area decreasing at when the radius is  15 cm?

Solution 

The volume of a sphere is  V = %284%2F3%29%2Api%2Ar%5E3.


Since both the volume and the radius depend on time, the formula becomes  V(t) = %284%2F3%29%2Api%2Ar%5E3%28t%29.


The derivative over time is  %28dV%29%2F%28dt%29 = 4%2Api%2Ar%5E2%28t%29%2A%28%28dr%29%2F%28dt%29%29


Hence,  %28%28dr%29%2F%28dt%29%29 = %28%28dV%29%2F%28dt%29%29%2F%284%2Api%2Ar%5E2%28t%29%29 = 120%2F%284%2Api%2A15%5E2%29.   (1)


The surface area of a sphere is  A = 4%2Api%2Ar%5E2,  or  A(t) = 4%2Api%2Ar%5E2%28t%29.


The derivative over time is  %28%28dA%29%2F%28dt%29%29 = 4%2Api%2A2%2Ar%28t%29%2A%28%28dr%29%2F%28dt%29%29 = 8%2Api%2Ar%28t%29%2A%28%28dr%29%2F%28dt%29%29


Substitute here  the expression  %28%28dr%29%2F%28dt%29%29 = 120%2F%284%2Api%2A15%5E2%29  from (1)  and r(t) = 15 cm.  You will get


    %28%28dA%29%2F%28dt%29%29 = 8%2Api%2A15%2A%28120%2F%284%2Api%2A15%5E2%29%29 = %288%2A15%2A120%29%2F%284%2A15%5E2%29 = %282%2A120%29%2F15 = 2*8 = 16 cm^2/min.

Problem 2

A circular plate is cooled and contracts so its radius is shrinking by  0.01 cm/s.
How fast is the area of one face changing when the radius is  2.5 cm?

Solution 

The area of the plate is  A = pi%2Ar%5E2,

Since the radius depends on time  r = r(t),  the area is


    A(t) = pi%2Ar%5E2%28t%29.      (1)


The derivative A'(t) is


    A'(t) = 2*pi*r(t)*r'(t).     (2)


We are given  r'(t) = 0.01 sm/s  and  r(t) = 2.5 cm.  Substitute these values into the formula (2) and get the


ANSWER.  A'(t) = 2*pi*2.5*0.01 = 2*3.14159*2.5*0.01 = 0.157 cm^2/s.

Problem 3

Water runs into a conical tank at the rate of  8 cubic meters per hour.
If the height of the cone is  10  meters,  and the diameter of its opening is  12 meters,
how fast is the water level rising when the water is  3 meters deep?

Solution 

The part of the conical tank, occupied by water, is the cone with the ratio radius to the height
of  r%2Fh = 6%2F10 = 0.6.


Hence, the volume of the water in the tank at every time moment t is

    V(t) = %281%2F3%29%2Api%2Ar%5E2%2Ah = %281%2F3%29%2Api%2A%280.6%2Ah%29%5E2%2Ah = 0.12%2Api%2Ah%5E3.    (1)


Differentiate it by the time

    %28dV%29%2F%28dt%29 = 0.12%2Api%2A3h%5E2%2A%28%28dh%29%2F%28dt%29%29.    (2)


%28dV%29%2F%28dt%29 is given: it is the inflow rate, or 8 m^3/hour.  
Hence, when h = 3 meters deep, we have from (2)


    8 = 0.12%2Api%2A3%2A3%5E2%2A%28%28dh%29%2F%28dt%29%29 = 0.12%2A27%2Api%2A%28%28dh%29%2F%28dt%29%29.


Hence,  %28dh%29%2F%28dt%29 = 8%2F%280.12%2A27%2A3.14159265%29 = 0.78595 meters per hour.


Rounding to 3 decimals, you get the 


ANSWER.  When the water depth is 3 meters, the water rising rate is about 0.786 m/hour.

Problem 4

Water flows into a tank having the form of frustum of a right circular cone.
The tank is  4 m tall with upper radius of  1.5 m and the lower radius of  1 m.
When the water in the tank is  1.2 m deep,  the surface rises at the rate of  0.012 m/s.
Calculate the inflow rate of water into the tank in  m3/s.

Solution

Here's how to calculate the discharge of water flowing into the tank,  step by step.

                The geometry connections 

We have the great cone with the base radius of 1.5 m and the height H m  (H is measured from the vertex).

We have the small cone with the base radius of 1 m   and the height (H-4) m.  This cone is cut out.

Therefore, for the small cone we have this proportion from similarity

     1.5%2FH = 1%2F%28H-4%29,  which gives  1.5H- 1.5*4 = H,  1.5H - H = 6,  0.5H = 6,  H = 6/0.5 = 12.

So, the great cone has the height H = 12 m;  the small cone has the height 12-4 = 8 m,  
and for every current h from vertex and r we have 

    r%2Fh = 1.5%2F12 = 0.5%2F4 = 0.125.


Therefore, for any current h and r,  r = 0.125h, where h is measured from the vertex.

In particular, when r = 1 m (the lower radius),  h = 12-4 = 8 meters from the cone vertex.

When the height of the water in the tank is 1.2 meters, h = 8 + 1.2 = 9.2 meters and  r = 0.125*(8+1.2) = 1.15 m.

                The volume of the water in the tank 

When the level of the water is h meters from the vertex, the volume of the water is

    V = %28pi%2F3%29%2Ar%5E2%2Ah - %28pi%2F3%29%2A1%5E2%2A8%5E2 m^3.


Substituting here  r = 0.125h, we have this formula for the volume of the water in the tank

    V = %28pi%2F3%29%2A0.125%5E2%2Ah%5E3 - %28pi%2F3%29%2A64 m^3.


We can consider here V and h as functions of time

    V(t) = %28pi%2F3%29%2A0.125%5E2%2A%28h%28t%29%29%5E3 - %28pi%2F3%29%2A64 m^3.    (1)

                Differentiate the volume of the water in the tank 

Write the derivative of the volume of the water in the tank in respect to the time.
The constant term in formula (1) does not matter, so forget about it.

    %28dV%29%2Fdt%29 = %28pi%2F3%29%2A0.125%5E2%2A3h%5E2%2A%28%28dh%29%2F%28dt%29%29,

Apply it for h = 8+1.2 = 9.2 meters.  It will give

    %28dV%29%2F%28dt%29 = %283.14159%2F3%29%2A0.125%5E2%2A3%2A9.2%5E2%2A%28%28dh%29%2F%28dt%29%29 = 4.154752775%2A%28%28dh%29%2F%28dt%29%29


Now substitute the given value of the rate of the surface rise  %28dh%29%2F%28dt%29 = 0.012 m/s.  You will get this equation

    %28%28dV%29%2F%28dt%29%29 = 4.154752775%2A0.012 = 0.049857  (rounded).


ANSWER.  The inflow rate into the tank is about 0.049857 m^3/s.

Problem 5

At noon,  ship  A  is  130 km  west of ship  B.  Ship  A  is sailing east at  30 km/h
and ship B is sailing north at  25 km/h.  How fast is the distance between the ships changing at  4:00 PM?

Solution 

In a coordinate plane (x.y), ship A and ship B are in the coordinate line "west - east" initially.

In standard notations, this coordinate line is horizontal  y = 0.


For ship A, its initial coordinate is (-130,0).

For ship B, its initial coordinate is (0,0).



In 4 hours, at 4:00 pm, ship A is at the point (-10,0),
                  while ship B is at the point (0,100).


The distance between the ships at t= 4:pm is

    D0 = sqrt%28%28-10%29%5E2%2B100%5E2%29 = sqrt%28100%2B10000%29 = sqrt%2810100%29.



The parametric form of the path for ship A is (-10+30t,0);
                                for ship B    (0, 100+25t),
where 't' is the time after 4:00 pm.


The square of the distance between the ships in parametric form is

    D^2(t) = (-10+30t)^2 + (100+25t)^2.


Take the time derivative  

    2%2AD0%2A%28%28dD%29%2F%28dt%29%29 = 2*(-10)*30 + 2*100*25.


Simplify

    2%2AD0%2A%28%28dD%29%2F%28dt%29%29 = 4400


and find the rate of the distance change between the ships

    %28dD%29%2F%28dt%29 = 4400%2F%282%2AD0%29 = 2200%2Fsqrt%2810100%29 = 21.891 km/h  (rounded).    ANSWER


At this point, the solution is complete.

Problem 6

A plane flies horizontally at an altitude of  6 km and passes directly over a tracking telescope on the ground.
When the angle of elevation is  π/3,  this angle is decreasing at a rate of  π/3 rad/min.
How fast is the plane traveling at that time?

Solution 

We have a horizontal line  y = 0  representing the ground surface.

We also have a horizontal line  y = 6 km  representing the trajectory of the plane.


The telescope (the observer) is point T on the ground.

We have point V on the line  y = 6 km  representing the plane when it is directly over an observer,
so the line TV is perpendicular to the ground surface y = 0.


Let point P on the line y = 6 represents the current position of the plane P = P(t).


The line VP is the trajectory of the plane, and triangle TVP is a right-angled triangle
with angle VPT = alpha = alpha%28t%29  which represents the elevation angle.


The length of the leg  |VT| = L(t)  is the covered distance, and the derivative  %28dL%28t%29%29%2F%28dt%29 
is the speed of the plane along the line y = 6 km.


We can write  tan%28alpha%29 = 6%2FL,  or  L = L(t) = 6%2Ftan%28alpha%28t%29%29.


Take the time derivative, considering L(t) as a composite function.  You will  get

    %28dL%28t%29%29%2F%28dt%29 = d%2F%28dt%29  %286%2Ftan%28alpha%28t%29%29%29 = d%2F%28dt%29 %28%286%2Acos%28alpha%28t%29%29%29%2Fsin%28alpha%28t%29%29%29%29) = %28-6%29%2Fsin%5E2%28alpha%28t%29%29 * alpha' .    (1)


We are given that  at the time moment t  alpha%28t%29 = pi%2F3 radians  and  alpha' = -pi%2F3 radians per minute.


We substitute these values into formula (1),  and we get the speed of the plane

    %28dL%28t%29%29%2F%28dt%29 = [ %28-6%29%2Fsin%5E2%28pi%2F3%29 ] * [ -pi%2F3 ] = [ %28-6%29%2F%28sqrt%283%29%2F2%29%5E2 ] * [ -pi%2F3 ] = (%286%2A4%29%2F3) * %28pi%2F3%29 = 8%2A%283.14159%2F3%29 = 8.3776 kilometers per minute,

    or  8.3776*60 = 502.65 kilometers per hour.     ANSWER


At this point, the problem is solved completely.

Problem 7

A rocket is being launched vertically over a point  A  on the ground with
a velocity of  550 𝑚𝑖/ℎ𝑟.  Twenty five miles away from point  A  on the ground,
there is a photographer video-taping the launch.  At what rate is the angle of elevation of the camera
changing when the rocket achieves an altitude of  25 miles ?

Solution 

Let line y = 0 in coordinate plane (x,y) represents ground surface, which we suppose to be horizontal.


Let A = (0,0) be the starting point of the rocket at the zero ground level 
   and let point  R = (0,y) represents the current position of the rocket at the height of y miles over the ground.


Let P = (25,0) be the position of the photographer at the ground level 25 miles from the start.


Thus we have a right-angled triangle ARP with the legs  |AR| = 25 miles and |AP| = 25 miles.


Let  theta  be the angle of elevation of the rocket to the horizontal line y = 0.


We are given  dy/dt = 550 mi/hr (rocket's velocity).


We want to find  dθ/dt  when y = 25 miles.


Angle theta  and  coordinate  'y' are related 

   tan%28theta%29 = y%2F25.


Differentiate both sides of this equation with respect to time (t):

    sec^2(θ)*(dθ/dt) = %281%2F25%29%2A%28%28dy%29%2F%28dt%29%29.


Solve for dθ/dt

    dθ/dt = %28cos%5E2%28theta%29%2F25%29%2A%28%28dy%29%2F%28dt%29%29



Find  cos%28theta%29  when  y = 25 miles.

When y = 25 miles, the triangle is a right isosceles triangle, so  theta = 45 degrees or π/4 radians.  
Therefore,  cos%28theta%29 = cos(45°) = 1%2Fsqrt%282%29.


Substitute and calculate:

    dθ/dt = %281%2F%28sqrt%282%29%5E2%29%29%2A%281%2F25%29+%2A+550,

    dθ/dt = %281%2F50%29%2A550 = 11 radians/hr.


    +-----------------------------------------------------------------------------+
    |    Conversion from mi/hr of the left side to radians/hr in the right side   |
    |               is just made inside the previous formula.                     |
    |         This conversion is already built into the coefficients.             |
    +-----------------------------------------------------------------------------+


ANSWER.  The angle of elevation is changing at a rate of 11 radians per hour, or 10.50423  degrees per minute, when the rocket reaches an altitude of 25 miles.


My other lessons on Calculus word problems at this site are
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    - Find the derivative of a function defined by complicated expression
    - Taking derivative of a function, which is defined implicitly
    - Find the derivative for a function satisfying given functional equation
    - Find the range of  f(x) = (5*cos(x))/(x + 1)),  x >=0
    - A tricky Calculus problem on derivative and anti-derivative
    - Couple of non-standard Calculus problems
    - Finding the minimum of a function defined on a curve in a coordinate plane
    - Tricky solution to find the maximum of a function defined by a complicated expression
    - Maximize the area of a trapezoid
    - Maximize the volume of an open box
    - Minimize surface area of a rectangular box with the given volume
    - Minimize the cost of an aquarium with the given volume
    - Minimize surface area of a conical paper cup with the given volume
    - Find the volume of a solid obtained by rotation of some plane shape about an axis
    - Finding the volume of a solid body mentally
    - OVERVIEW of my lessons on Calculus word problems

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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