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The subject of mathematics is related to almost all the other subjects. The advancement in the fields of engineering, science, economics statistics etc. are facilitated by use of mathematics. In other words, the application of mathematics helps in development and easier understanding of topics in other subjects. The branch of mathematics that is used for such a purpose is called as applied mathematics. It may be noted that in this branch of math, the important terms and constants used in the other topics also form as parts.
Let us give a simple introduction to applied mathematics with interesting examples from other fields.
A car leaves an airport at 8 am and runs at an average velocity 45 mph. At 9 am another car leaves the same airport in the same direction and it has to meet the first car before noon. What should be the minimum average velocity with which the second car should run to achieve the necessity?
The situation described in the above physics problem is not very unusual. A person traveling in the first car might have left out something and his friend at the airport may like to reach that article. And by noon, the first person might reach his destination and may not be reachable thereafter. Let us see how applied mathematics helps us to solve. The velocity refers to the rate of change of distance with respect to time. Hence the distance traveled by the first car in the 4 hours (from 8 am to noon) is given by the mathematical equation d1 = v1* t1 = 45*4 = 180 miles, since we know that v1 = 45 mph and t1 = 4 hours. The concept of applied math is same for the second car but now the equation is d2 = v2* t2. But in this case the known values are d2 = 180 miles, since at the point of interception both cars must have traveled the same distance and out of necessity the maximum value of t2 can only be 3 hours (from 9 am to noon). Therefore, 180 = v2*3, which gives the solution as v2 = 60 mph.
Let us study another problem related to physics but which can be considered as an applied mathematics. Two wires of ½ in. diameter are anchored at a ceiling roof as shown in the diagram. These wires are riveted to a hook which is used to hold heavy weights. The wires make angles of 45o and 60o with the ceiling. The wires have an ultimate tensile strength of 16T per sq.in. What could be the maximum weight that can be loaded on the hook?
The concept of this problem is used in material lifting equipment. Ultimate tensile strength of 16T/ sq.in. means the wire can take a load of only 16T for a cross section area of 1 in. Since the diameter of the wires are ½ in. each wire can take only a load of 16(π/4)(1/2)2 = π tons ≈ 3.14 tons. Now mathematically we can draw a vector diagram and find the solution. The same is drawn below.
The wires on the left and right take the loads that are the projection of the main load W in the direction of wires. Let those components be P and Q respectively. As per vector algebra, P = (√2)W/2 and Q = (√3)W/2. Obviously the magnitude of Q is greater and therefore it must be equal to 3.14T. Hence W can be equal to a maximum of 3.14/0.866 = 3.63T approximately.
The concept of matrices is widely used in statistical fields. We will give an introduction to matrices in our next topic.