﻿ Reasons for change. Exercises and examples solved step by step

# Reasons for change. Exercises and examples solved step by step

The problems of reasons for change are to calculate how a given magnitude, known as the variation of another magnitude related to it and the value of those magnitudes at a given time, will vary over time.

How much does one magnitude A increase or decrease over time if we know the variation with respect to time of another magnitude B?

The derivatives and allow us to calculate the variation of magnitudes with respect to time.

One of the difficulties of these problems lies in finding the formula that relates the magnitudes involved.

On the other hand, it should be borne in mind that changing magnitudes are no longer fixed values, but are time-dependent functions, since they have a different value for each instant of time.

Let’s look at it with several examples.

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## Change ratio problems solved

### Exercise 1

In a circle, we know that its radius increases at a rate of 1 cm/s. What is the ratio of change of the area of the circle when the radius is equal to 5 cm? In this problem we are being told that the ratio of change of radius is 1 cm/s. The reason for change of a magnitude is its derivation with respect to time, therefore: They are asking us the ratio of change of the area of the circumference when r=5 cm, i.e. the ratio of the area to the time: In other words, we are asked how much the area will be growing when the radius is equal to 5 cm.

Now we have to find a formula that relates the area to the radius of the circle, which we have in the formula of the area of a circle: As I mentioned earlier, both the area and the radius are not constant values, but are time-dependent functions.

To find the variations of each magnitude over time, we derive in both members of the equation with respect to time and it remains: In the first member, we have derived A with respect to t, whose derivative is dA/dt.

In the second, to derive r with respect to t, we use the chain rule (from the outside in): the r² derivative is 2r and we multiply it by the r derivative that is dr/dt.

Once we have derived, we replace the data that the statement gives us:

r=5 cm

dr/dt=1 cm/s In this expression we can already calculate dA/dt.

The radius is in cm and the ratio of the radius change is in cm/s. We operate with units in mind: And we will have the result in cm²/s ### Exercise 2

The volume of a cube is changing at the rate of 75 cm³/minute.

a) Find the reason for change on your side when you are 5 cm tall

b) Find the rate of change of the surface area when it is 24 cm².

Paragraph a:

We know that the volume changes at the rate of 75 cubic centimeters per minute: And they ask us for the reason for change on their side when they are two inches tall: The formula that relates the volume to the “a” side of the cube is: We derive in both members of the equation: And we substitute dV/dt and a for their values: From where we can clear da/dt: And there it is: Paragraph b:

As in the previous section, the volume changes at the rate of 75 cubic centimetres per minute: And this time they ask us the reason for the change of the surface area when it is 24 cm²: The formula that relates the area of the cube to the “a” side of the cube is: Deriving with respect to time on both sides of the equation, we have left: On this occasion we have no data directly from to or from da/dt.

We know that the instant we want to calculate da/dt, the area is equal to 24, therefore, in the previous equation, substituting A for 24, we can obtain the value of a: We still have to obtain the value of da/dt, as we have the value of a, we will calculate it as in the previous section, from the formula of the volomen: We derive with respect to time in both members of equality: And we substitute dV/dt and a for their values: We clear da/dt: And we calculate its value: Now we have all the data to calculate dA/dt, so we can replace it in the equation where we derive the area and the side of the cube: When we replace it, it’s up to us: So dA/dt will be: ### Exercise 3

A worker holds a rope 36 m long and at the other end there is a weight. The rope passes through a pulley located at a height of 20 meters. If it moves away from the pulley at a rate of 5 m/s, at what speed does the weight rise when it is 10 meters above the original position?

At the beginning, the outline of the problem would be as follows: The worker moves away from the pulley at a rate of 5 m/s, when z=10 m therefore: On the one hand, the statement tells us that the length of the rope is 36 m. The string corresponds to the sides z e and the triangle, therefore: On the other hand, by Pythagoras, relate the three sides: Since we want to relate the magnitude z (which is the distance that the weight moves) to the magnitude x, which is the distance that the worker moves, of the first equation, we can clear the y in function of z: And replace this expression of and in the expression of Pythagoras: Now we drift to both sides of the same time: We have all the data except the value of x.

From the expression obtained from Pythagoras, we substitute z for 10 and we will have an expression that only depends on x, from where we can obtain its value: And now we have all the values to clear dz/dt: We cleared it: And we finally figured it out: 