Navigation
 Chapter 1  Fundamentals
 Chapter 2  Algebraic Functions

Chapter 3  Applications
 Curvature and Radius of Curvature
 Maxima and Minima  Applications

Time Rates  Applications
 0102 Water flowing into cylindrical tank
 03 Water flowing into rectangular trough
 0405 Water flowing into triangular trough
 0607 Ladder slides down the wall
 0809 Rate of movement of shadow on the ground
 10  A boy on a bike
 1112 Two trains; one going to east, and the other is heading north
 1314 Water flowing into trapezoidal trough
 1516 Movement of shadow from light at eye level
 1718 Rate of shadow in the wall of a building
 1921 Two cars driving in parallel roads
 2224 One car from a city starts north, another car from nearby city starts east
 25 Two cars that may collide at 12:30 PM
 2627 Time Rates: Kite moving horizontally
 2829 Time Rates: Two cars driving on roads that intersects at 60 degree
 30  Two trains in perpendicular railroad tracks
 3132 Train in an elevated track and car in perpendicular road
 3334 Time Rates: A car traveling east and airplane traveling north
 3536 Time Rates: Lengthening of shadow and movement of its tip in 3D space
 3738 How fast a ship leaving from its starting point
 Rate of change of surface area of sphere
 Chapter 4  Trigonometric and Inverse Trigonometric Functions
 Partial Derivatives
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 400000=120[14π(D2−10000)]
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I believe the answer should
I believe the answer should be negative since the surface area is decreasing
It is not necessary to…
In reply to I believe the answer should by Anonymous (not verified)
It is not necessary to indicate a negative to the rate of change because the problem already said that it is decreasing.
However, the negative is necessary if the problem asks for "How fast is the surface area is changing". For this statement, it is important to determine if the rate of change is decreasing or increasing or perhaps it is an integral part of the problem. The negative sign is necessary for such a case, or an absolute value with word decreasing or increasing will also do.