Question

Find the equation of the curve whose slope at any point is 𝑑𝑦/𝑑𝑥 = 𝑥^2 √𝑥 and which passes through the point (1,0).

Answer 1

The slope of the curve is given by $\frac{dy}{dx}=x^2.\sqrt{x}$ .....(1)

Integrating both side of (1) with respect to $x$ ,we get

$\intop dy=\intop x^2.\sqrt{x} dx$

$\implies \intop dy=\intop x^{\frac{5}{2}} dx$

$\implies y=\frac{x^{(\frac{5}{2}+1)}}{\frac{5}{2}+1}+C$

$\implies y=\frac{2}{7}.x^{\frac{7}{2}}+C$ [where $C$ is an integrating constant]

As the curve passes through the point $(1,0)$ we have,

$0=\frac{2}{7}.1^{\frac{7}{2}}+C$

$\implies C=-\frac{2}{7}$

Therefore the required equation of the curve is $y=\frac{2}{7}.x^{\frac{7}{2}}-\frac{2}{7}$