# Tutor profile: Arun B.

## Questions

### Subject: Calculus

FInd $$\lim_{x\to 1} \frac{x^{20}-1}{x^{12}-1}$$

If we apply the value of limit to the function, then we get $$\frac{0}{0}$$ which is not the correct solution. $$\therefore$$ we can use the following trick to solve the problem. Divide both numerator and denominator by $$x-1$$ $$lim_{x\to 1} \frac{x^{20}-1}{x^{12}-1}= lim_{x\to 1} \frac{ \frac{x^{20}-1}{x-1}}{ \frac{x^{12}-1}{x-1}}$$ $$lim_{x\to 1} \frac{x^{20}-1}{x-1}\div lim_{x\to 1} \frac{x^{12}-1}{x-1}$$ Now we can use the theorem, $$lim_{x\to a} \frac{x^{n}-a^n}{x-a}= nxa^{n-1}$$ $$\rightarrow lim_{x\to 1} \frac{x^{20}-1}{x-1}=20x$$ and $$ lim_{x\to 1} \frac{x^{12}-1}{x-1}=12x$$ $$\therefore lim_{x\to 1} \frac{x^{20}-1}{x^{12}-1}=\frac{20}{12}=\frac{5}{3}$$

### Subject: Physics

A car and a bike both move on a highway at a speed of 54 Km/Hr. The car is ahead of the motorcycle by 300m. The biker decides to overtake the car. What should be his acceleration if he wants to overtake the car in one minute?

The initial velocity of the car and bike are the same. $$ U_{car}=U_{bike}=54 Km/Hr = 54 *\frac{5}{18} m/s= 15 m/s$$ In one minute the car travels a distance of $$ U_{car}*60=15*60=900 m$$ In order to overtake the car the bike has to cover a total distance of 900+300=1200m in one minute. Using equation of motion $$ S_{bike}=U_{bike} *t+\frac{1}{2}*a_{bike}*t^2$$ we can find the acceleration of the car Substituting the values we get, $$1200=15*60+\frac{1}{2}*a_{bike}*60^2$$ Solving for $$a_{bike}$$ we get, $$a_{bike}=\frac{1}{6} m/s^2$$

### Subject: Differential Equations

Differential equations are all around us. In-depth analysis and differential equations are essential in solving complex modern-day problems. It has application in physics, chemistry, biology, economics etc. Let us understand the application of differential equations in finance. Q) The principal grows steadily at 10% per year in a bank. Calculate in how many years $100 double itself?

Let P be the principal at any point in time. From the question, $$\frac{dP}{dt} = (\frac{10}{100})*P$$ $$\rightarrow \frac{ dP}{dt}=\frac{P}{10}$$ Using variable separation we can write, $$\frac{dP}{P}=\frac{dt}{10}$$ Integrating both sides we get, $$log P=\frac{t}{10}+C_1$$ Where $$C_1$$ is the constant of integration Taking exponential on both sides we get, $$P=e^{\frac{t}{10}} . e^{C_1} $$ Let $$e^{C_1}=C$$, hence $$P=C*e^{\frac{t}{10}} $$ At $$t=0$$ $$P=100$$ (Inital condition) $$\therefore P=100*e^{\frac{t}{10}}$$. Let as assume in $$t_1$$ time $$P$$ become $200 $$\therefore 200=100*e^{\frac{t_1}{10}}$$ or $$t_1=10 log (2)$$

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