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Let y=f(x)y=f(x) be a curve passing through (1,1)(1,1) such that the triangle formed by the cordinate axis and the tangent at any point of the curve lies in the first quadrant and has area 22. Form the differential equation and determine all such curves

#Calculus

Note by Akhilesh Prasad
5 years, 8 months ago

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Comments

Let P(x0,f(x0))P(x_{0}, f(x_{0})) be any point of f(x)f(x) and the tangent line through PP be represented by:

y=f′(x0)x+[f(x0)−x0f′(x0)] y = f'(x_{0})x + [f(x_{0}) -x_{0}f'(x_{0})] (i)

with O(0,0)O(0,0) as the origin and Q(0,f(x0)−x0f′(x0)),R(x0f′(x0)−f(x0)f′(x0),0)Q(0, f(x_{0}) -x_{0}f'(x_{0})), R(\frac{x_{0}f'(x_{0}) - f(x_{0})}{f'(x_{0})}, 0) as the yy and xx intercepts respectively. If right triangle QORQOR has a constant area of 22, then the area is computed according to:

12[f(x0)−x0f′(x0)][x0f′(x0)−f(x0)f′(x0)]=2\frac{1}{2} [f(x_{0}) - x_{0}f'(x_{0})][\frac{x_{0}f'(x_{0}) - f(x_{0})}{f'(x_{0})} ] = 2 (ii)

From (ii) we now deduce:

f(x0)−x0f′(x0)=A,x0f′(x0)−f(x0)f′(x0)=4Af(x_{0}) - x_{0}f'(x_{0}) = A, \frac{x_{0}f'(x_{0}) - f(x_{0})}{f'(x_{0})} = \frac{4}{A}

such that f′(x0)=−A24,f(1)=1⇒f(x0)=−A24x0+B⇒f(x0)=−A24x0+(1+A24) f'(x_{0}) = -\frac{A^2}{4}, f(1)=1 \Rightarrow f(x_{0}) = -\frac{A^2}{4} x_{0} + B \Rightarrow f(x_{0}) = -\frac{A^2}{4} x_{0} + (1 + \frac{A^2}{4}) (iii)

Substitution of (iii) into (ii) yields A2=4A^2 = 4 and ultimately f(x0=−x0+2\boxed{f(x_{0} = -x_{0} + 2} .

tom engelsman - 1 year ago
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