Please help me on a problem about particular antiderivatives.

Sep 2018
27
0
Philippines
Hello everyone. Our professor discussed this particular problem the other day:

At any point (x,y) on a curve, d2y/dx2 = 1 - x^2. The equation of the tangent line to the curve at (1,1) is y = -x + 2. Find the equation of the curve.

I have notes on the solution but I’m stuck. I don’t understand why dy/dx equals to -1. Can someone please lead me in the right direction? Attached is the picture of my lecture notes.

Thank you very much!
 

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romsek

MHF Helper
Nov 2013
6,665
3,002
California
It looks like you correctly derived $\dfrac{dy}{dx}$, now just integrate it.

$\dfrac{dy}{dx} = -1$ at $x=1$ because you selected the constant of integration, $-\dfrac 5 3$ to make it so.
 
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Plato

MHF Helper
Aug 2006
22,470
8,640
Hello everyone. Our professor discussed this particular problem the other day:
At any point (x,y) on a curve, d2y/dx2 = 1 - x^2. The equation of the tangent line to the curve at (1,1) is y = -x + 2. Find the equation of the curve.
I have notes on the solution but I’m stuck. I don’t understand why dy/dx equals to -1. Can someone please lead me in the right direction? Attached is the picture of my lecture notes.
Sorry but I cannot read your images.
But based upon the post, if $\dfrac{d^2y}{dx^2}=1-x^2$ then $\dfrac{dx}{dy}=x-\tfrac{1}{3}x^3+C$ Now surely you know that $\dfrac{dx}{dy}$ is slope.
And we are given that $y=-1x+2$ is the tangent at $(1,1)$ which has slope $-1$.
Thus $\dfrac{dx}{dy}(1,1)=1-\tfrac{1}{3}1^3+C=-1$ so $C=~?$

The find the primitive(anti-derivative) again. Use the point $(1,1)$ to find the constant in the primitive.
 
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Sep 2018
27
0
Philippines
Thanks everyone! So it was the slope all along! :D