what is the rate of change from x = π to x = 3 pi over 2

Problem 1

A rectangular water tank (see effigy beneath) is being filled at the constant rate of 20 liters / 2d. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of modify of the top of water in the tank?(express the answer in cm / sec).

Solution to Problem i:

• The volume V of water in the tank is given by.
  V = w*L*H
  
• Nosotros know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. Five and H are functions of time. We tin can differentiate both side of the above formula to obtain
  dV/dt = W*L*dH/dt
  
• note W and L do not change with time and are therefore considered as constants in the above performance of differentiation.
  
• We now find a formula for dH/dt as follows.
  dH/dt = dV/dt / W*L
  
• We need to convert liters into cubic cm and meters into cm as follows
  1 litter = ane cubic decimeter
  = chiliad cubic centimeters
  = m cm ⁱⁱⁱ
  and 1 meter = 100 centimeter.
  
• Nosotros now evaluate the rate of change of the meridian H of water.
  dH/dt = dV/dt / W*Fifty
  = ( 20*one thousand cm ³ / sec ) / (100 cm * 200 cm)
  = 1 cm / sec.

Trouble ii

An plane is flying in a straight management and at a abiding height of 5000 meters (see effigy below). The angle of elevation of the airplane from a fixed point of observation is a. The speed of the airplane is 500 km / 60 minutes. What is the rate of alter of angle a when it is 25 degrees? (Express the answer in degrees / second and round to one decimal place).

Solution to Problem 2:

• The airplane is flight horizontally at the rate of dx/dt = 500 km/hour. We need a relationship between bending a and distance x. From trigonometry, we tin can write
  tan a = h/x
  
• angle a and altitude ten are both functions of fourth dimension t. Differentiate both sides of the above formula with respect to t.
  d(tan a)/dt = d(h/ten)/dt
  
• We at present utilise the chain rule to further aggrandize the terms in the above formula
  d(tan a)/dt = (sec ² a) da/dt
  d(h/x)/dt = h*(-1 / x ²) dx/dt.
  (note: height h is abiding)
  
• Substitute the above into the original formula to obtain
  (sec ² a) da/dt = h*(-i / x ²) dx/dt
  
• The above tin can be written equally
  da/dt = [ h*(-1 / x ²) dx/dt ] / (sec ^two a)
  
• We now use the first formula to find 10 in terms of a and h follows
  x = h / tan a
  
• Substitute the above into the formula for da/dt and simplify
  da/dt = [ h*(- tan ² a / h ²) dx/dt ] / (sec ² a)
  = [ (- tan ²a / h) dx/dt ] / (sec ^two a)
  = (- sin ²a / h) dx/dt
• Utilise the values for a, h and dx/dt to estimate da/dt with the right conversion of units: 1km = grand m and 1 hour = 3600 sec.
  da/dt = [- sin ^two(25 deg)/5000 yard]*[500 000 thou/3600 sec]
  = -0.005 radians/sec
  = -0.005 * [ 180 degrees / Pi radians] /sec
  = -0.3 degrees/sec

Problem 3

If two resistors with resistances R1 and R2 are continued in parallel as shown in the effigy beneath, their electric beliefs is equivalent to a resistor of resistance R such that

ane / R = 1 / R1 + 1 / R2

If R1 changes with time at a rate r = dR1/dt and R2 is abiding, express the rate of change dR / dt of the resistance of R in terms of dR1/dt, R1 and R2.

Solution to Trouble 3:

• Nosotros start by differentiating, with respect to fourth dimension, both sides of the given formula for resistance R, noting that R2 is constant and d(ane/R2)/dt = 0
  (-1/R ²)dR/dt = (-1/R1 ^two)dR1/dt
  
• Arrange the above to obtain
  dR/dt = (R/R1) ²dR1/dt
  
• From the formula 1 / R = 1 / R1 + ane / R2, we can write
  R = R1*R2 / (R1 + R2)
  
• Substitute R in the formula for dR/dt and simplify
  dR/dt = (R1*R2 / R1*(R1 + R2)) ²dR1/dt
  = (R2 / (R1 + R2)) ²dR1/dt

Exercises

ane - Find a formula for the rate of modify dV/dt of the volume of a balloon being inflated such that it radius R increases at a charge per unit equal to dR/dt.
2 - Find a formula for the rate of modify dA/dt of the area A of a foursquare whose side x centimeters changes at a rate equal to 2 cm/sec.
3 - Ii cars starting time moving from the same bespeak in 2 directions that makes ninety degrees at the constant speeds of s1 and s2. Find a formula for the rate of change of the altitude D between the ii cars.

Solutions to the Above Exercises

1 -    dV/dt = 4*Pi*R ² dR/dt
two -    dA/dt = 4x cm ² /sec
3 -    dD/dt = sqrt( s1 ² + s2 ² )
More references on calculus problems

fosterthats1949.blogspot.com

Source: https://www.analyzemath.com/calculus/Problems/rate_change.html

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