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The line tangent to a curve, no derivative

  • Thread starter icantadd
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Homework Statement


Let (x0,y0) be a point on the graph f(x) = x2 for x0 not equal to 0. Find the equation of the line tangent to the graph of f at that point by finding a line that intersects the cure in exactly one point. Do not use the derivative to find this line.


Homework Equations



f(x) = x2
y=mx+b
(y1-y2) = m(x1-x2)


The Attempt at a Solution



The way I would normally approach this problem is to use the derivative. But that is a syntax error in this problem; however, it does not stop me from knowing that the slope should depend on x0, where the slope of some x0 should be 2*x0.

At the same time, any point on the graph can be given in the form (x0,(x0)2). I tried to force through on this point, and get an equation (y- (x0)2) = m(x-x0). I would think I need to solve for m, but then there are too many variables, and I get too many possible solutions.

Also, I tried to set x2 = m*x + b, to see the set of solutions. That is unhelpful, there are solutions to that set of equations that are not tangents to the graph, but instead are secants.

I believe that my general approach to the problem is wrong; I am missing something. I would love any advice.
 

Answers and Replies

  • #2
Dick
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Your second approach is the right one. The curve is y=x^2, the line is y=m*(x-x0)+x0^2 (by forcing it to go through the right point as you did in the first part). So set x^2=m*(x-x0)+x0^2. Solve for x. For a general value of m, you will get two solutions (one, of course, is x=x0). What's the other one? Now find the value of m where those two solutions are equal.
 
  • #3
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Thank you sir.

Playing with the algebra starting with x^2 = m*(x-x0) +x0^2, I get:

(x-x0)(x+x0)=m*(x-x0), which simplifies to

m = x + x0 <=> x = m -x0. The other solution of x is x = x0 given by Dick.

Setting the two equal to each other is:

m - x0 = x0 Therefore,

m = 2*x0.

Which is what the answer should be. Brilliant, thank you again.

Question: To get that one solution is x = x0, did you notice that both the left and right hand sides of the equation in
(x-x0)(x+x0) = m*(x-x0) are equal if both are 0, which happens when x=x0, or is there another way to get that?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
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Thank you sir.

Playing with the algebra starting with x^2 = m*(x-x0) +x0^2, I get:

(x-x0)(x+x0)=m*(x-x0), which simplifies to

m = x + x0 <=> x = m -x0. The other solution of x is x = x0 given by Dick.

Setting the two equal to each other is:

m - x0 = x0 Therefore,

m = 2*x0.

Which is what the answer should be. Brilliant, thank you again.

Question: To get that one solution is x = x0, did you notice that both the left and right hand sides of the equation in
(x-x0)(x+x0) = m*(x-x0) are equal if both are 0, which happens when x=x0, or is there another way to get that?
That, and that you deliberately set the tangent equation up so that at x=x0, the two would be equal.
 
  • #5
HallsofIvy
Science Advisor
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This was, by the way, Fermat's method for finding tangents.
 

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