Thread: Find equation of a curve

1. [Solved]Find equation of a curve

Been working on revision and still waiting on my mate to bring me the notes, so in the meantime I wouldn't like stalling, anyways this is the question

Find $\displaystyle \int{\frac{1}{x^2}dx}$

The gradient of a curve is given by $\displaystyle \frac{dy}{dx}=\frac{1}{x^2}$. Find the equation of the curve, given that it passes through the point (1,3).

2. Originally Posted by Lonehwolf
Been working on revision and still waiting on my mate to bring me the notes, so in the meantime I wouldn't like stalling, anyways this is the question

Find $\displaystyle \int{\frac{1}{x^2}dx}$

The gradient of a curve is given by $\displaystyle \frac{dy}{dx}=\frac{1}{x^2}$. Find the equation of the curve, given that it passes through the point (1,3).
$\displaystyle \int{x^{a}dx}=\frac{x^{a+1}}{a+1} +C ~$

$\displaystyle \int{\frac{1}{x^2}dx}=~\int{x^{-2}dx}=\frac{x^{-2+1}}{-2+1} +C$

Thus equation of such curves =$\displaystyle \frac{x^{-1}}{-1} +C$

This curve passes through (1,3)

So $\displaystyle 3 = \frac{(1)^{-1}}{-1} +C$

$\displaystyle 3 = -1 +C \qquad \implies C= 4$

$\displaystyle \frac{-1}{x} +4$ is the requirement