The following exercises try to calculate the variation rate of a quantity when the variation rate of another quantity related to it is known. In this type of exercise, the "variation rate" is interpreted as a derivative and, in most cases, it is enough to use the chain rule to obtain what is requested. You have to choose the units according to the data of the problem; For example, if a volume is measured in liters, we will have to measure lengths with decimeters.
Exercise 1. How fast does the level of the water contained in a cylindrical tank drop if we are emptying it at a rate of 3000 liters per minute?
Solution
Let r be the radius of the cylinder and h be the height measured in decimeters. Let V(t) be the volume of water, measured in liters (=dcm3), that is in the cylinder at time t measured in minutes. The information they give us is a variation rate
V(t + 1) − V(t) = −3000 liters per minute
In this type of exercise, the variation rate is interpreted as a derivative: V ′(t) = −3000. Notice that V(t + to) − V(to) u V′(to)t, so the interpretation is reasonable. The negative sign of the derivative is required since the volume decreases with time. Since the radius is constant but the height of the water depends on time, we have
and we deduce therefore
V(t) = πr2h(t)
V′(t) = −3000 = πr2h′(t)
h′(t) = −3000 decimeters per minute πr2
If we express the measurements in meters, then h ′ (t) = − 3 meters per minute. πr2
Exercise 2. A point P moves on the part of the parabola x = y 2 situated in the first quadrant in such a way that its x coordinate is increasing at a rate of 5 cm/sg. Find the speed at which point P recedes from the origin when x = 9.
Miguel Martín and Javier Pérez (University of Granada)
Solved exercises chapter 2 2 Solution
Let (x(t), y(t)) be the coordinates, measured in centimeters, of the point P at the instant t measured in seconds. They tell us that y(t) > 0 and that x(t) = y(t)2. The distance from point P to the origin is given by f (t) = px(t)2 + y(t)2 , so
′ x(t)x ′(t) + y(t)y ′(t) f (t) = px(t)2 + y(t)2
What we are asked for is f ′(to) knowing that x(to) = 9. In this case it must be y(to) = 3. We also know x′(t) = 5 (cm/sg). With this it is easy to deduce the value of y′(to) = x′(to) = 5. Finally,
2y(to) 6
′ x(to)x′(to)+y(to)y′(to) 45+3(5/6) 95
f (to)= px(to)2+y(to)2 = 81+9 =6√10 cm/s sorry is a lot