A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Carissa15

  • one year ago

Please help. I have a calculus question about the secant of a parabola y=x^2 that passes through points A=(2,4) and A'=(2+delta x, 4+ delta y) , find the slope of this secant. I am then given a) delta x =0.01 and b) delta x = E. To answer this do I just sub in the a) and b) values of x and solve the equation?

  • This Question is Closed
  1. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hint: the slope m of your secan t, is: \[m = \frac{{{\delta _y}}}{{{\delta _x}}}\]

  2. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    furthermore, we can write this: \[\Large 4 + {\delta _y} = {\left( {2 + {\delta _x}} \right)^2}\]

  3. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you then solve for delta y? As values for delta x are already given?

  4. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    from my last equation, we have: \[\Large 4 + {\delta _y} = 4 + {\left( {{\delta _x}} \right)^2} + 4{\delta _x}\] so we can simplify as below: \[\Large \frac{{{\delta _y}}}{{{\delta _x}}} = {\delta _x} + 4\]

  5. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1438251569667:dw|

  7. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, that makes sense. I do not understand how the two parts of a) delta x=0.01 and b) delta x=E come into this? Sorry, probably very silly question, does not get covered in my textbook and getting confused.

  8. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    delta x= E means delta x= infinity?

  9. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, epsilon

  10. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think taht we have to substitute: \[\Large \begin{gathered} {\delta _x} = 0, \hfill \\ {\delta _x} = \varepsilon \hfill \\ \end{gathered} \] into the formula above

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think you want to finish part \(a\) first let \(\delta_x=0.01\) and find the slope of secant

  12. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oops..: \[\Large \begin{gathered} {\delta _x} = 0.01, \hfill \\ {\delta _x} = \varepsilon \hfill \\ \end{gathered} \]

  13. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Great, once the formula is broken down. I have now substituted delta x=0.01 and found delta y is =0.0401. Will do the same with Epsilon (except won't get a numerical value)

  14. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    knew what you meant :)

  15. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[4+\delta y=4+(\delta x)^2+4 \delta x\] then subtracted 4 on both sides to isolate \[\delta y\]

  16. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    That looks good!

  17. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then got \[( \delta x)^2+4 \delta x\] then sub in my x value for a)

  18. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    For part \(a\), below works too : \[m= \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{(2+0.01)^2-2^2}{0.01}\]

  19. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks

  20. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    am i correct in thinking that with epsilon I will not be able to find a value, I will just be left with an equation after substitution?

  21. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1438252734043:dw|

  22. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes, plugin \(\delta_x = \epsilon\) and simplify the secant expression as much as you can

  23. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    great, thank you so much. So much clearer now :)

  24. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    np, for part \(b\) i think you should get : \[m= \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{(2+\epsilon)^2-2^2}{\epsilon} = \dfrac{4\epsilon +\epsilon^2}{\epsilon} = 4+\epsilon\]

  25. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Looks your teacher wants you see how a secant line approaches to tangent line by decreasing the value of \(\epsilon\)

  26. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    As simplified as I can? is that what you mean?

  27. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes, but I have simplified that for you already

  28. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    when using \[ \frac{ \delta y }{ \delta x }\] I get a different result than y2-y1/x2-x1

  29. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    both should give you same answer, show me ur work

  30. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\delta y = (\delta x)^2+4 \delta x\] then after substitution i get \[\delta y = (0.01)^2+4(0.01) = 0.0001+0.04 = 0.0401\]

  31. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and with \[\frac{ y2 -y1}{ x2-x1 }=\frac{ (2+0.01)^2-2^2}{ 0.01 } =4.01\]

  32. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    looks fine, whats wrong with that ?

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    notice that slope of secant equals \(\dfrac{\delta y}{\delta x}\), not just \(\delta y\)

  34. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok, so my first one i still need to divide by delta x. then they will be the same. :)

  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yep! Easy, slope = (change in y)/(change in x)

  36. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thats the highschool definition of slope, and it still works! :)

  37. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Awesome, thank you so much. Lifesaver!

  38. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Here is a little animation that shows how the secant line approaches/diverges from tangent line as \(\delta_x\) is decreased/increased |dw:1438254466089:dw|

  39. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \(A\) is a fixed point. Notice that as the point \(A'\) approaches the point \(A\), the secant line approaches the tangent line.

  40. Carissa15
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Great, thanks!

  41. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    np

  42. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.