# Mathematics to solve real-world problem

# Analysing problem of global warming using Mathematics

we studied mathematics in high-school, degree college but never tried to apply the concepts in real-world. I was just thinking what example I shall take and how to write a program for the same. I got to know an interesting problem, only the problem is interesting but in real the problem is really worrying.

**Problem – WRITING SOME WHAT DIFFERENT PROGRAM, SIMPLE BUT DIFFERENT**

in Arctic the **ice surface reflects 90% of the radiation** but the **water surface reflects only 10% of the radiation**. Considering the fact in last two decades the Arctic like places showing effect of climate change. The water surface absorbs 90% of the radiation and hence leads to chain reaction. Assuming for every 1000 gray radiation, 0.65 sqft/hr water formation from ice surface, to predict how much ice surface might vanish and what could be the impact on global warming in next two decades?

To solve this problem one must apply integration which we study in Mathematics calculus, the same problem in other way to find out the rate of change of water surface can be evaluated using differential calculus.

Let me try out this problem with a C program, definitely I still didn’t find the integration or differential calculus equation to solve, but let me prepare set of simulated data with certain assumptions.

**C Program**

#include <stdio.h> #include <stdlib.h> #include <time.h> /* Ice surface - reflects 90% of the sun light Water surface - reflects 10% of the sub light 100 ice surface - 10% absorbs 100 water surface - 90% Radiation = 100 gray 100 * 90 = 9000 gray 100 * 10 = 1000 gray assume that as temperature decreases ice surface increases, at the rate say 0.6 sqft per 1000 gray */ struct ICE_RADIATION_INFO{ float iceSurfaceArea; float waterSurfaceArea; float radiationQuantity; float iceSurfaceAbsorption; float waterSurfaceAbsorption; float waterFormationPerHour; }; typedef struct ICE_RADIATION_INFO ICE_RADIATION_INFO; int main(int argc, const char * argv[]) { ICE_RADIATION_INFO iceRadiationSamples[100]; short int index; // Just to initialise the set of data for analysis - for (index = 0; index < 100; ++index) { ICE_RADIATION_INFO *iceRadiationSample = &iceRadiationSamples[index]; // 1.5 is constant taken to generate area in floating point number iceRadiationSample->iceSurfaceArea = (rand() % 100) * 1.5; iceRadiationSample->waterSurfaceArea = iceRadiationSample->iceSurfaceArea; iceRadiationSample->radiationQuantity = (rand() % 1000) * 1.5; // absorption is as per discussion 90% from water surface and 10% from ice surface iceRadiationSample->iceSurfaceAbsorption = iceRadiationSample->radiationQuantity * 0.1; iceRadiationSample->waterSurfaceAbsorption = iceRadiationSample->radiationQuantity * 0.9; // with an assumption for every 1000 gray radiation, 0.65 sqft water forms iceRadiationSample->waterFormationPerHour = ( 0.65 * iceRadiationSample->radiationQuantity) / 1000; // 0.65 sqft/hr for 1000 gray /* Every hour 0.65 sqft water occumulation , this intern absorbs 90% of radiation , here have to apply integration dx/dt = 0.65 */ } printf("Surface Area \t Radiation \t\t Ice Absorption \t Water absorption\n"); for (index = 0; index < 50; ++index) { ICE_RADIATION_INFO iceRadiationSample = iceRadiationSamples[index]; printf("%f\t\t %f \t\t %f \t\t %f\n",iceRadiationSample.iceSurfaceArea,iceRadiationSample.radiationQuantity,iceRadiationSample.iceSurfaceAbsorption,iceRadiationSample.waterSurfaceAbsorption); } return 0; }

OUTPUT –

**Surface Area Radiation Ice Absorption Water absorption**

**10.500000 373.500000 37.349998 336.149994**

**109.500000 987.000000 98.699997 888.299988**

**45.000000 408.000000 40.799999 367.200012**

**66.000000 1317.000000 131.699997 1185.300049**

**34.500000 1063.500000 106.349998 957.150024**