Finding Solutions for Function F(x): Perpendicular Tangents & Slopes

In summary: I see. Thank you very much for all of your help!In summary, the conversation discusses finding the coordinates of the point of intersection of two perpendicular tangent lines to a function F(x) = kx^2 + 3, as well as verifying the solutions for parts a, b, and c of the problem. The experts suggest graphing the solutions and choosing a specific value for k to better understand the problem. It is also mentioned to explain why choosing either the positive or negative root for t will give the same result for the tangent lines. Lastly, "w.l.og" is explained to stand for "without loss of generality."
  • #1
carlodelmundo
133
0
Hi All.

Homework Statement



Consider the function F(x) = kx^2 + 3.

a.) If the tangent lines to the graph of F at (t, F(t) ) and (-t , F(-t)) are perpendicular, find t in terms of k.

b) Find the slopes of the tangent lines mentioned in part (a).

c.) Find the coordinates of the point of intersection of the tangent lines mentioned in part (a).


Homework Equations



1.) First Derivative
2.) Algebraic Manipulation

The Attempt at a Solution



I need to verify my answers. I HAVE SCANNED ALL OF MY ANSWERS SO YOU CAN READ IT EASIER. http://carlodm.com/calc/prob1.jpg .[/URL]

Here are my questions:


For a.) I got t = +- sqrt( 1 / 4k^2)). I chose the positive root but is this correct? I mean, I chose the positive root to make my calculations simpler but is there a good reason why to choose the positive, or negative?

For b.) Is this algebraic manipulation legitimate?

For c.) This seems a little weird for me considering its not a "finite" answer. Does this answer work? Thank you!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


Regarding your answer for a), fix k (and therefore t) see what positive and negative answers give you. Try graphing them.

Now take a look at b), you basically have 2 tangent lines L1 and L2, if you pick t = 1/(2k) then the slope of L1 is 1 and slope of L2 is -1. Now what would have happened if you chose t = -1/(2k)?

For c) your answer is very much finite, it's just a function of k, why does that seem weird to you? Your original function did not specify a k therefore your tangent lines will be dependent on k as well, as will the point of their intersesction.
 
Last edited:
  • #3


NoMoreExams,

Thanks for the quick reply. I'm dumbfounded by your explanation in a.). How can I "Fix k?" Is there an Algebra error? Or are you trying to say to graph t in terms of k and look at the graph?

If it's the latter, I graphed that solution curve. It looks like a diamond shaped curve, but I don't know what I can deduce from this curve.

Fot b.)... I understand that if t is the negative root...then the direction of the tangent lines will be opposite (negative slope of tangent will become the positive slope).

Thank you.
 
  • #4


By fix k I mean pick a value for k, say... k = 1, now graph your function and your 2 lines. Your function shouldn't look diamond shaped.
 
  • #5


I see. I let k = 5. so F(x) = 5x^2 + 3 and F'(x) = 10x.

When I finish the Algebra, I'm left with t = +- (1/10). Which is equivalent to what I found in my answer in a.).

The graph is two lines. This just proves that it's correct, right?
 
  • #6


Sure I meant plot both your two tangent lines AND your original function. Now that you fixed k = 5, pick the positive root i.e. t = 1/10 that means your first tangent line will have a point (1/10, F(1/10)) and your 2nd one will have the point (-1/10, F(-1/10)). What if you picked t = -1/10, what would have changed between 2 lines, would they have looked any different?
 
  • #7


No it does not look any different meaning choosing the positive or negative t will give the same results.

Should I explicitly state that we are "assuming" that t is the positive root for AP purposes?

Thanks for your help--for b) and c) are my answers logical?
 
  • #8


You should probably explain that choosing either will give the same result since you are just switching between the 2 lines I believe. Maybe someone else can confirm/deny.

Yep b) and c) look logical to me, did you understand what I said about c) though?

I am not sure what the standards are for AP exams so I can't tell you if what you have is "enough" or if you need to write more. You definitely should not write +/- 1/(2k) = 1/(2k) since that's simply not true; you COULD write w.l.og., let t = 1/(2k) or something of the sort and explain why.
 
  • #9


Thank you NoMoreExams! Part c.) Makes perfect sense.

I have one last question. What does "w.l.og" stand for?
 
  • #10


without loss of generality
 

1. What is a perpendicular tangent?

A perpendicular tangent is a line that intersects a curve at a right angle, or 90 degrees. It is also known as a normal line.

2. How do you find the slope of a perpendicular tangent?

The slope of a perpendicular tangent can be found by taking the negative reciprocal of the slope of the tangent line. For example, if the slope of the tangent line is 2, the slope of the perpendicular tangent will be -1/2.

3. What is the relationship between perpendicular tangents and parallel lines?

Perpendicular tangents have slopes that are negative reciprocals of each other, while parallel lines have the same slope. This means that perpendicular tangents will never be parallel to each other.

4. How do you find the point of intersection of two perpendicular tangents?

To find the point of intersection of two perpendicular tangents, set the equations of the tangents equal to each other and solve for the x and y values. This will give you the coordinates of the point of intersection.

5. Can there be more than one perpendicular tangent to a curve at a given point?

Yes, there can be more than one perpendicular tangent to a curve at a given point. This can happen when the slope of the curve is changing rapidly, creating multiple points where the slope of the tangent is the negative reciprocal.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
467
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
110
  • Calculus and Beyond Homework Help
Replies
3
Views
631
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
191
  • Calculus and Beyond Homework Help
Replies
4
Views
794
  • Calculus and Beyond Homework Help
Replies
1
Views
786
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top