Calculus proof involving slopes of tangents - help needed

  • #1
The tangent at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.


y=x^3, dy/dx=3x^2, y=mx+b


I tried setting the derivative equal to the original function to see when they intersect and go from there but then realized that the tangent at a certain point is not the same as the equation for the slope of the tangent at -any- point. I looked at this thread https://www.physicsforums.com/showthread.php?t=94520 but still could quite figure it out, I wasn't sure whether to post in a 5 year old thread or make a new one so feel free to move me if I made the wrong choice. If someone could get me started that would be great.
 

Answers and Replies

  • #2
Dick
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Don't post in old threads, ok? It's really annoying. I assume anyone that does that is just being too lazy to post their own. Pick a point on the curve (x0,x0^3). Then the tangent line is (y-x0^3)=(3*x0^2)*(x-x0), right? Intersect that with y=x^3 to find the other value of x where they intersect besides at x=x0. Hint: x^3-x0^3 is divisible by x-x0. It's not really a cubic equation.
 
  • #3
Ok, wasn't sure because I've been yelled at on forums for not using the search function before making a thread. Say I use the point P(a,a^3) (or x0's or whatever), I don't understand why the equation is (y-a^3)=(3a^2)(x-a). I would substitute the point into the equation for a line y=mx+b and end up with a^3=3a^2(x)+b. Could you explain why you are subtracting the point from the x and y value and you do not have a b value in your equation?
 
  • #4
Dick
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a^3=3a^2(x)+b isn't the equation of a line, where's y?? I'm using the line form (y-y0)=m*(x-x0). It's the point-slope form. I strongly disagree with the person that yelled at you for not searching the forums first. You certainly can do that, somebody may have posted the same problem before, but that doesn't mean their confusion about the same problem is the same as your confusion.
 
  • #5
In my equation, y is on the left side y=a^3 (for the point (a,a^3)) I was confused by the formatting of your equation, I don't think I've ever actually had to use it before. I see the connection now between it and the one I have used to find slope between 2 points (m=(y2-y1)/(x2-x1)). I'll go give that a try and see if it works out for me, thanks for the help. Just out of curiosity though, is there a way to do it using the equation y=mx+b?
 
  • #6
Dick
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If you happen to know the y-intercept b, then y=mx+b is handy. In this case you don't. You know a point on the line (a,a^3). So (y-a^3)=m*(x-a), is the way to go. You can convert that into y=mx+b by expanding and solving for b, but why bother, it's just making the problem harder.
 
  • #7
So I rearranged and set those two equations equal to one another, I end up with x^3=a^2(3x-2a), not sure where to go from there, I don't see any difference of cubes to factor like you mentioned earlier.
 
  • #8
Dick
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I would have thought it would lead you to (x^3-a^3)=(3*a^2)*(x-a). I don't think I went wrong. Where did you? Show your steps.
 
  • #9
y-a^3=3a^2(x-a)
y-a^3=3a^2(x)-3a^3
y=3a^2(x)-3a^3+a^3
y=3a^2x-2a^3
set equal to y=x^3
x^3=3a^2x-2a^3
x^3=a^2(3x-2a)

new step, probably no more useful... x^3=a^3(3x/a-2)
 
  • #10
Dick
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Ok, you decided to bypass my suggestion to factor x^3-a^3. Fine. Now factor x^3-3*a^2*x+2a^3=0. It can be done. Hint: and you'll probably ignore this as well, but one of the factors is (x-a). How did I know that?
 
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  • #11
Ok, after a failed attempt at long division, a failed attempt at synthetic division, a realization of my mistake, and then a successful attempt at synthetic division I end up with (x-a)(x^2+ax-2a^2)=0.
So using the quadratic formula on my second factor I got the roots x=a and x=-2a.
So, they intersect when x=-2a.... so the slope at Q is 12x^2 which is 4 times the slope at P which is 3x^2. Sweet, thanks a lot.
 
  • #12
Dick
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Right. I don't even think the first thread you were looking at was very good. Better to solve it on your own. Don't you think?
 
  • #13
I very much agree.
 

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