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.

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?

reason for change

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:

reason for change examples

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: problems with the ratio of change

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:

change reason exercises

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:

reason for change exercises

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

reason for change resolved exercises

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:

examples of reason for change

And we will have the result in cm²/s

reasons for change

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:

average change ratio resolved exercises

And they ask us for the reason for change on their side when they are two inches tall:

ratio of change

The formula that relates the volume to the “a” side of the cube is:

change reason exercises solved for secondary

We derive in both members of the equation:

exercises reason for change

And we substitute dV/dt and a for their values:

change reason problems solved

From where we can clear da/dt:

reason for change solved examples

And there it is:

example of reason for change

Paragraph b:

As in the previous section, the volume changes at the rate of 75 cubic centimetres per minute:

reason for change derived

And this time they ask us the reason for the change of the surface area when it is 24 cm²:

related change reasons resolved exercises

The formula that relates the area of the cube to the “a” side of the cube is:

derived change ratios

Deriving with respect to time on both sides of the equation, we have left:

change ratio exercises

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:

exercises resolved for reason of change

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:

change ratio problems

We derive with respect to time in both members of equality:

reasons for change examples

And we substitute dV/dt and a for their values:

reason for instant change resolved exercises

We clear da/dt:

how the rate of change is calculated

And we calculate its value:

change ratio exercises

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:

reason for change derived from resolved exercises

When we replace it, it’s up to us:

problems reason for change

So dA/dt will be:

reason for change differential calculation

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:

foro de clase razón de cambio

The worker moves away from the pulley at a rate of 5 m/s, when z=10 m therefore:

reason for change calculation

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:

problems with the reason for change

On the other hand, by Pythagoras, relate the three sides:

related reasons resolved exercises

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:

foro de clase: razón de cambio

And replace this expression of and in the expression of Pythagoras:

change ratio problems

Now we drift to both sides of the same time:

reason for change solved problems

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:

derived as reason for change

And now we have all the values to clear dz/dt:

change reason exercises resolved

We cleared it:

change reason exercise

And we finally figured it out:

examples of reasons for change