Carissa15
  • Carissa15
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?
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
hint: the slope m of your secan t, is: \[m = \frac{{{\delta _y}}}{{{\delta _x}}}\]
Michele_Laino
  • Michele_Laino
furthermore, we can write this: \[\Large 4 + {\delta _y} = {\left( {2 + {\delta _x}} \right)^2}\]
Carissa15
  • Carissa15
you then solve for delta y? As values for delta x are already given?

Looking for something else?

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

More answers

Michele_Laino
  • 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: \[\Large \frac{{{\delta _y}}}{{{\delta _x}}} = {\delta _x} + 4\]
Carissa15
  • Carissa15
hmm
ganeshie8
  • ganeshie8
|dw:1438251569667:dw|
Carissa15
  • 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.
Michele_Laino
  • Michele_Laino
delta x= E means delta x= infinity?
Carissa15
  • Carissa15
sorry, epsilon
Michele_Laino
  • 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
ganeshie8
  • ganeshie8
I think you want to finish part \(a\) first let \(\delta_x=0.01\) and find the slope of secant
Michele_Laino
  • Michele_Laino
oops..: \[\Large \begin{gathered} {\delta _x} = 0.01, \hfill \\ {\delta _x} = \varepsilon \hfill \\ \end{gathered} \]
Carissa15
  • 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)
Carissa15
  • Carissa15
knew what you meant :)
Carissa15
  • Carissa15
\[4+\delta y=4+(\delta x)^2+4 \delta x\] then subtracted 4 on both sides to isolate \[\delta y\]
ganeshie8
  • ganeshie8
That looks good!
Carissa15
  • Carissa15
then got \[( \delta x)^2+4 \delta x\] then sub in my x value for a)
ganeshie8
  • ganeshie8
For part \(a\), below works too : \[m= \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{(2+0.01)^2-2^2}{0.01}\]
Carissa15
  • Carissa15
thanks
Carissa15
  • 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?
ganeshie8
  • ganeshie8
|dw:1438252734043:dw|
ganeshie8
  • ganeshie8
Yes, plugin \(\delta_x = \epsilon\) and simplify the secant expression as much as you can
Carissa15
  • Carissa15
great, thank you so much. So much clearer now :)
ganeshie8
  • ganeshie8
np, for part \(b\) i think you should get : \[m= \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{(2+\epsilon)^2-2^2}{\epsilon} = \dfrac{4\epsilon +\epsilon^2}{\epsilon} = 4+\epsilon\]
ganeshie8
  • ganeshie8
Looks your teacher wants you see how a secant line approaches to tangent line by decreasing the value of \(\epsilon\)
Carissa15
  • Carissa15
As simplified as I can? is that what you mean?
ganeshie8
  • ganeshie8
Yes, but I have simplified that for you already
Carissa15
  • Carissa15
when using \[ \frac{ \delta y }{ \delta x }\] I get a different result than y2-y1/x2-x1
ganeshie8
  • ganeshie8
both should give you same answer, show me ur work
Carissa15
  • Carissa15
\[\delta y = (\delta x)^2+4 \delta x\] then after substitution i get \[\delta y = (0.01)^2+4(0.01) = 0.0001+0.04 = 0.0401\]
Carissa15
  • Carissa15
and with \[\frac{ y2 -y1}{ x2-x1 }=\frac{ (2+0.01)^2-2^2}{ 0.01 } =4.01\]
ganeshie8
  • ganeshie8
looks fine, whats wrong with that ?
ganeshie8
  • ganeshie8
notice that slope of secant equals \(\dfrac{\delta y}{\delta x}\), not just \(\delta y\)
Carissa15
  • Carissa15
ok, so my first one i still need to divide by delta x. then they will be the same. :)
ganeshie8
  • ganeshie8
Yep! Easy, slope = (change in y)/(change in x)
ganeshie8
  • ganeshie8
thats the highschool definition of slope, and it still works! :)
Carissa15
  • Carissa15
Awesome, thank you so much. Lifesaver!
ganeshie8
  • 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|
ganeshie8
  • ganeshie8
\(A\) is a fixed point. Notice that as the point \(A'\) approaches the point \(A\), the secant line approaches the tangent line.
Carissa15
  • Carissa15
Great, thanks!
ganeshie8
  • ganeshie8
np

Looking for something else?

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