## 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?

1. Michele_Laino

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

2. Michele_Laino

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

3. Carissa15

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

4. Michele_Laino

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

hmm

6. ganeshie8

|dw:1438251569667:dw|

7. Carissa15

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

delta x= E means delta x= infinity?

9. Carissa15

sorry, epsilon

10. Michele_Laino

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

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

12. Michele_Laino

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

13. Carissa15

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

knew what you meant :)

15. Carissa15

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

16. ganeshie8

That looks good!

17. Carissa15

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

18. ganeshie8

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

thanks

20. Carissa15

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

|dw:1438252734043:dw|

22. ganeshie8

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

23. Carissa15

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

24. ganeshie8

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

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

26. Carissa15

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

27. ganeshie8

Yes, but I have simplified that for you already

28. Carissa15

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

29. ganeshie8

both should give you same answer, show me ur work

30. Carissa15

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

31. Carissa15

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

32. ganeshie8

looks fine, whats wrong with that ?

33. ganeshie8

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

34. Carissa15

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

35. ganeshie8

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

36. ganeshie8

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

37. Carissa15

Awesome, thank you so much. Lifesaver!

38. ganeshie8

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

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

40. Carissa15

Great, thanks!

41. ganeshie8

np