# Tutor profile: Akshat K.

## Questions

### Subject: Pre-Calculus

A right triangle has base $$x$$ and height $$y$$. If the area of the triangle is 5 square inches, express the perimeter of the triangle as a function of $$x$$

Using the Pythagorean Theorem, we know that the third side (hypotenuse) $$= \sqrt{x^2+y^2}$$ So, the perimeter $$P$$ will be: $$P=x+y+\sqrt{x^2+y^2}$$ Since $$A=\frac{1}{2}xy=5$$, $$=> xy=10$$ $$=> y=10/x$$ Now, Substituting back gives $$P=x+\frac{10}{x}+\sqrt{x^2+(10/x)^2}$$

### Subject: Calculus

$$\int_{}{} e^x sinx$$ $$dx$$

1) Use Integration by parts: $$ u = sinx$$ and $$dv = e^x$$ So, $$du = cosx$$ and $$ v = e^x$$ Now, $$\int_{}{} udv = uv - \int{}{}vdu$$ $$ = e^xsinx - \int{}{}e^xcosx $$ $$dx$$ Here, performing integration by barts on $$\int{}{}e^xcosx$$ $$dx$$ we get: $$ u = cosx$$ and $$dv = e^x$$ So, $$du = -sinx$$ and $$ v = e^x$$ Therefore, $$\int_{}{} e^x sinx$$ $$dx$$ $$= e^xsinx - [e^xcosx - \int{}{}e^x(-sinx)$$ $$dx$$] $$\int_{}{} e^x sinx$$ $$dx$$ $$=e^xsinx - e^xcosx - \int{}{}e^xsinx$$ $$dx$$ $$\int_{}{} e^x sinx$$ $$dx$$ $$ + \int{}{}e^xsinx$$ $$dx$$ $$=e^xsinx - e^xcosx + C$$ $$2\int_{}{} e^x sinx$$ $$dx$$ $$=e^xsinx - e^xcosx + C$$ $$\int_{}{} e^x sinx$$ $$dx$$ $$=\frac{e^xsinx - e^xcosx}{2} + C$$ $$\int_{}{} e^x sinx$$ $$dx$$ $$=\frac{e^x}{2}(sinx - cosx) + C$$

### Subject: Algebra

The population of a city is $$ P = 2500 e^{x^2 - x -6} $$ where $$x= 0$$ represents the population in 2000. In which year will the population be $$2500$$?

Given: $$ P = 2500$$; To find: $$x$$ Now, we know that: $$ P = 2500 e^{x^2 - x -6} $$ $$ 2500 = 2500 e^{x^2 - x -6} $$ $$ e^{x^2 - x -6} = \frac{2500}{2500}$$ $$e^{x^2 - x -6} = 1$$ $$e^{x^2 - x -6} = e^0$$ $$x^2 - x -6 = 0$$ $$(x-3)(x+2) = 0$$ $$ x = 3$$ or, $$x = -2$$ Now, since x represents number of years after 2000, it cannot be negative. Hence, $$x = 3$$ So, in the year 2003, the population will be 2500.

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