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?

- Carissa15

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- schrodinger

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- Michele_Laino

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

- Michele_Laino

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

- Carissa15

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

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## More answers

- 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

hmm

- ganeshie8

|dw:1438251569667:dw|

- 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

delta x= E means delta x= infinity?

- Carissa15

sorry, epsilon

- 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

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

- Michele_Laino

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

- 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

knew what you meant :)

- Carissa15

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

- ganeshie8

That looks good!

- Carissa15

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

- 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

thanks

- 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

|dw:1438252734043:dw|

- ganeshie8

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

- Carissa15

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

- 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

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

- Carissa15

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

- ganeshie8

Yes, but I have simplified that for you already

- Carissa15

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

- ganeshie8

both should give you same answer, show me ur work

- 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

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

- ganeshie8

looks fine, whats wrong with that ?

- ganeshie8

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

- Carissa15

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

- ganeshie8

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

- ganeshie8

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

- Carissa15

Awesome, thank you so much. Lifesaver!

- 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

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

- Carissa15

Great, thanks!

- ganeshie8

np

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