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Mathematics 10 Online
OpenStudy (anonymous):

Find an equation of the tangent line y = f(x) at x = a. f(x) = sin 4x, a = pi/8

zepdrix (zepdrix):

Hmm I think you've done a few of these types of problems. So I'm just curious, which result makes more sense to you. Seeing our tangent line in slope-intercept form: \(\large y=mx+b\), Or in point-slope form: \(\large y-y_o=m(x-x_o)\).

OpenStudy (anonymous):

in slope intercept from

zepdrix (zepdrix):

Ok we will have 2 pieces we need to find. \(\large m\) is the slope of our tangent line. Take the derivative of your function, and evaluate it at \(\large x=\pi/8\). That will be your \(\large m\) value.

OpenStudy (anonymous):

okay hold on

OpenStudy (anonymous):

when i find the derivative a = pi/8 is not int he equation right?

OpenStudy (anonymous):

the*

OpenStudy (anonymous):

i got sin(4)

zepdrix (zepdrix):

Take the derivative of sin(4x) first. Don't plug a=pi/8 in until after you've taken a derivative.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

i got 4 cos(4x)

zepdrix (zepdrix):

ok good, now plug in x=pi/8 :D

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i got 0 after i plugged in and solved

zepdrix (zepdrix):

Ok sounds good. So we've determined that the slope \(\large m\) of our tangent line will be 0. \[\large y=mx+b \qquad \rightarrow \qquad y=0x+b\] Now we just need to find the \(\large b\) value.

zepdrix (zepdrix):

To find it, we'll need to plug in a coordinate pair that falls on the line. Well, since our line is tangent to the curve at x=pi/8, we can plug x=pi/8 into that function to get a corresponding y value. This will be the coordinate pair we'll plug into our tangent line.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so i should plug in pi/8 into y = mx + b?

zepdrix (zepdrix):

no, plug it into your original function, before you took the derivative of it.

OpenStudy (anonymous):

ohh okay

OpenStudy (anonymous):

so sin 4(pi/8) ?

OpenStudy (anonymous):

okay so i got 1

zepdrix (zepdrix):

So that gives us a coordinate pair of \(\large \left(\dfrac{\pi}{8},\;1\right)\). Plug that into your tangent function to solve for \(\large b\).

OpenStudy (anonymous):

ok so y = 0x + b so i should plug it in place of b?

zepdrix (zepdrix):

No, your coordinate pair is \(\large (x,\;y)\). Plug it into those.

OpenStudy (anonymous):

ohh ok got it

OpenStudy (anonymous):

okay so 1 = 0(pi/8) + b

zepdrix (zepdrix):

So what is your b value? :3

OpenStudy (anonymous):

Lol sorry i got 1 :)

OpenStudy (anonymous):

b = 1

zepdrix (zepdrix):

Ok sounds good. So if we plug that missing \(\large b\) into our tangent line equation, we get,\[\large y=0x+1\]Which can be simplified to,\[\large y=1\]

OpenStudy (anonymous):

okay

zepdrix (zepdrix):

I guess it's really 3 steps. We had to find ~\(\large m\) ~\(\large b\) ~A Coordinate Pair.

zepdrix (zepdrix):

It's not too difficult, just a lot of silly letters :) Keep practicing!

OpenStudy (anonymous):

okay so our equation is complete? and yes just lot to do and thanks

zepdrix (zepdrix):

Yes, that's our final answer.

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