Integral Calculus: Finding the equation of the curve.

Submitted by Rexel Reedus on

Hello, can someone help me solve this problem? I tried working it around but I can't arrive at the correct answer. Thank you!

Problem: Find the equation of the curve for which y''=12/x3 if it passes through (1,0) and is tangent to the line 6x+y=6 at that point.

Answer: xy+6x=6

Source: Elements of Calculus and Analytic Geometry by Reyes and Chua

Tags
Equation of the curve

Jhun Vert Tue, 07/21/2015 - 07:07

$6x + y = 6$

$6 + y' = 0$

$y' = -6$
 

$y''=12x^{-3}$

$y' = -6x^{-2} + C_1$   →   Eq (1)

$y = 6x^{-1} + C_1x + C_2$   →   Eq (2)
 

At the point of tangency (1, 0), y' = -6

From Eq (1)
$-6 = -6 + C_1$

$C_1 = 0$
 

From Eq (2)
$y = 6x^{-1} + C_2$

$0 = 6 + C_2$

$C_2 = -6$

Thus,
$y = \dfrac{6}{x} - 6$   →   Eq (2)

$xy = 6 - 6x$

$xy + 6x = 6$       answer