Verification requested

• Jan 14th 2011, 08:45 PM
Foxlion
Verification requested
Let D(n) denote change in some integer n; let A(x) and A(y) denote either the corresponding x or y co-ordinate at some point A on the euclidean plane;

Question: Considering two points P and Q upon the curve y= x^2-2x-3 of the euclidean plane what is the slope of the secant line PQ at the fixed point P as Q approaches P?

slope m of secant line PQ = D(y)/D(x)

so, if D(y)= Q(y) - P(y) then Q(y)= P(y) + D(y)

and if D(x)=Q(x) - P(x) then Q(x)=P(x) + D(x)

i.e. D(y) = [x^2 + 2xD(x) + D(x)^2 - 2x - 2D(x) - 3] - [x^2 - 2x - 3]

and furthermore D(y)/D(x)= 2x + D(x) - 2

am uncertain of final solution and would like to know if this answer implies the limit of the slope of the secant line as D(x) approaches 0 is 2x - 2
• Jan 15th 2011, 08:13 AM
HallsofIvy
Quote:

Originally Posted by Foxlion
Let D(n) denote change in some integer n; let A(x) and A(y) denote either the corresponding x or y co-ordinate at some point A on the euclidean plane;

Question: Considering two points P and Q upon the curve y= x^2-2x-3 of the euclidean plane what is the slope of the secant line PQ at the fixed point P as Q approaches P?

slope m of secant line PQ = D(y)/D(x)

so, if D(y)= Q(y) - P(y) then Q(y)= P(y) + D(y)

and if D(x)=Q(x) - P(x) then Q(x)=P(x) + D(x)

i.e. D(y) = [x^2 + 2xD(x) + D(x)^2 - 2x - 2D(x) - 3] - [x^2 - 2x - 3]

and furthermore D(y)/D(x)= 2x + D(x) - 2

am uncertain of final solution and would like to know if this answer implies the limit of the slope of the secant line as D(x) approaches 0 is 2x - 2

The problem asked what happens "as Q approaches P", which means, yes, D(x) goes to 0 and the the slope of the secant line goes to 2x- 2. That, by the way, implies that the equation of the tangent line, at \$\displaystyle (x_0, y_0)\$ is \$\displaystyle y= (2x_0- 2)(x- x_0)+ y_0\$.

I presume that this is the very beginning of a Calculus or Analysis course and you will so be introduced to the "derivative" as "slope of the tangent line".
• Jan 15th 2011, 08:56 PM
Foxlion
Quote:

Originally Posted by HallsofIvy
I presume that this is the very beginning of a Calculus or Analysis course and you will so be introduced to the "derivative" as "slope of the tangent line".

you presume correct, thank you.