Answer to Question #294828 in Calculus for Silvius

Question

At any point (x, y) on a curve, 𝑑^2𝑦/𝑑𝑥^2 = 1-x^2

, and an equation of the tangent line to the

curve at the point (1, 1) is y = 2 – x. Find an equation of the curve.

Answer 1

Solution:

$\dfrac{d^2y}{dx^2}=1-x^2 \\\Rightarrow \dfrac{dy}{dx}=x-\dfrac{x^3}3+C$

Given, tangent $y = 2 – x$

So, slope of tangent $=m=-1$

At (1, 1), $m=\dfrac{dy}{dx}$

$\Rightarrow x-\dfrac{x^3}3+C=m \\\Rightarrow 1-\dfrac13+C=-1 \\ \Rightarrow C=-\dfrac53$

So, $\dfrac{dy}{dx}=x-\dfrac{x^3}3-\dfrac53$

Also, at any point, $m=\dfrac{dy}{dx}$

$x-\dfrac{x^3}3-\dfrac53=-1 \\ \Rightarrow x=-2,x=1$

For $x=-2: y=2-x=2+2=4$

For $x=1:y=2-x=2-1=1$

Points are (-2,4), (1,1).

Consider $\dfrac{dy}{dx}=x-\dfrac{x^3}3-\dfrac53$ again.

On integrating:

$y=x^2-\dfrac{x^4}{12}-\dfrac53x+C_1$

Put (1,1) in this and solve for C₁

$1=1-\dfrac{1}{12}-\dfrac53+C_1 \\\Rightarrow C_1=\dfrac74 \\ \therefore y=x^2-\dfrac{x^4}{12}-\dfrac53x+\dfrac74$