logo

Ask Questions, Get Answers

X
 
Home  >>  CBSE XII  >>  Math  >>  Model Papers

The slope of the tangent of any point on a curve is \( \lambda \) times the slope of the line joining the point of contact to the origin. Formulate the differential equation and hence find the equation of the curve.

Download clay6 mobile app

1 Answer

Toolbox:
  • Slope of the tangent to the curve is $\large\frac{dy}{dx}$
  • If the linear differential equation is of the form $\large \frac{dy}{dx}$$=f(x)$ then it can be solved by seperating the variables and then integrating.
  • $\int\large\frac{dx}{x}=$$\log x$
Step 1:
Let $P(x,y)$ be any point on the given curve.
The slope of the tangent at P=$\lambda$(slope of line joining the point P and the origin)
(i.e) $\large\frac{dy}{dx}=$$\lambda\big(\large\frac{y-0}{x-0}\big)$
$\Rightarrow \large\frac{dy}{dx}=$$\lambda \large\frac{y}{x}$
This is the required differential equation.
$\large\frac{dy}{dx}=\frac{\lambda y}{x}$
Step 2:
Separating the variables we get,
$\large\frac{dy}{y}=$$\lambda \large\frac{dx}{x}$
Integrating on both sides we get,
$\int \large\frac{dy}{y}=$$\lambda\int\large\frac{dx}{x}$
(i.e)$\log y=\lambda \log x+\log C$
$\Rightarrow \log y=\log x^{\lambda}.C$
$y=Cx^{\lambda}$
This is the equation of the required curve.
answered Sep 30, 2013 by sreemathi.v
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
...
X