Question

Find an equation of the tangent line to the curve
y = sin(7 x) + cos(4 x)

at the point (pi/6,y(pi/6) ). Tangent line:

Answer 1

Hey Carol c:
So we're trying to find a line that is tangent to our curve at a specific point.
That line will be of the form:\[\Large\rm y=\color{orangered}{m}x+b\]Where m, the slope, is given by the derivative function evaluated at that particular point.

Answer 2

Yes I know I would plug in pi/6 to the equation but I keep getting a weird answer

Answer 3

\[\Large\rm y=\color{orangered}{f'\left(\frac{\pi}{6}\right)}x+b\]

Answer 4

\[\Large\rm f(x)=\sin(7x)+\cos(4x)\]Did you have any trouble taking this derivative?

Answer 5

(I'm calling it f(x) instead of y, otherwise we'll confuse it with our tangent line y D: )

Answer 6

It would be 7cos7x-4sin4x
correct?

Answer 7

mmm good good.

Answer 8

and then i plug in pi/6

Answer 9

and i get (-1/2)-(2radical3)

Answer 10

Hmm I'm coming up with:\[\Large\rm f'\left(\frac{\pi}{6}\right)=7\left(-\frac{\sqrt3}{2}\right)-4\left(\frac{\sqrt3}{2}\right)\]

Answer 11

Maybe I made an algebraic error. But what would my next step be?

Answer 12

\[\Large\rm \cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt3}{2}\]

Answer 13

That value is the slope of our tangent line.
Let's simplify it down before plugging it in.

Answer 14

So we have -7 of those things, and -4 more of them.
So we have -11sqrt3/2, something like that, yah?

Answer 15

yes thats correct

Answer 16

So here is what our tangent line currently looks like with our new found slope,
\[\Large\rm y=\color{orangered}{-\frac{11\sqrt3}{2}}x+b\]

Answer 17

that makes sense

Answer 18

To find b, the y-intercept, we need to be able to plug in a point that lies on the line.

Answer 19

Since the line is `tangent` or `touching` the curve at one specific point,
the curve and tangent line will share that coordinate point.

Answer 20

But notice that they're giving you a little bit of extra work, they only gave you the x coordinate :)

Answer 21

thats true

Answer 22

:(

Answer 23

So we need to find out what the coordinate point is that the two functions share.
We need the y coordinate that corresponds to x=pi/6.

So now you'll plug pi/6 into the `original function` to get a coordinate pair.

Answer 24

hmm

Answer 25

i get -1

Answer 26

mmm me too, ok good good.
So the point we can use for our tangent line is:\[\Large\rm \left(\frac{\pi}{6},-1\right)\]

Answer 27

yay!

Answer 28

So plug in the stuff c: what do you get for your b?
It might be a little messy looking!

Answer 29

is my b the -1?

Answer 30

So far we've establisshed this:\[\Large\rm \color{royalblue}{y}=-\frac{11\sqrt3}{2}\color{royalblue}{x}+b\]And we can use our coordinate point \(\Large\rm \left(\color{royalblue}{\dfrac{\pi}{6}},\color{royalblue}{-1}\right)\) to find our b.

Answer 31

ahhh i see so then...\[-1=\frac{ -11\sqrt{3} }{ 2 }(\pi/6)\]

Answer 32

With a +b on the end, yes.

Answer 33

yes
and thenn

Answer 34

im confused what to do with the pi/6

Answer 35

Multiply your fractions, bottom with bottom, top with top.
The you'll need to do some addition to isolate your b.

Answer 36

\[-b=-66\sqrt{3}*2\pi\]

Answer 37

and plus one at the end

Answer 38

Woah, b should have been a fraction. :o
Where did the denominator go?

Answer 39

I cross multiplied? I'm sorry that I'm not getting this as easy as it should be

Answer 40

When we multiply fractions, we multiply straight across, yes?
So this:\[\Large\rm -1=-\frac{11\sqrt3}{2}\cdot\frac{\pi}{6}+b\]Gives us this:
\[\Large\rm -1=-\frac{11\sqrt3 \pi}{12}+b\]From here we're NOT allowed to cross multiply since that +b is hanging out over there.

If both sides were completely fractions, then yes we could do so.

Answer 41

You want to add that ugly monstrous thing to the other side.

Answer 42

that makes much more sense

Answer 43

Do i have to simplify further after adding it to the other side or is that my final answer?

Answer 44

\[\Large\rm b=\frac{11\sqrt3~\pi}{12}-1\]You can get a common denominator if you prefer to look at your b as a single term.
But this is fine right here.

For calculus it's nice to get in the habit of `not oversimplifying things`.

Answer 45

Just make sure you realize that this is not your final answer, you need to plug this into your tangent line to get your final answer.

Answer 46

ok so then...

Answer 47

\[\frac{ -11\sqrt{3} }{ 2 }x+\frac{ 11 \sqrt{3}}{12 }\]

Answer 48

theres a pi
and -1

Answer 49

idk why it keeps cutting off

Answer 50

hehe

Answer 51

yah that sounds right! \c:/
\[\Large\rm y=-\frac{11\sqrt3}{2}x+\frac{11\sqrt3~\pi}{12}-1\]

Answer 52

Here is graph of the function so you can see what it looks like and confirm that this crazy mess is correct.
https://www.desmos.com/calculator/qrj741tptf

Answer 53

let me plug in to check right now

Answer 54

It's kind of hard to read since the slope is so steep,
but the blue line is tangent to the red curve at pi/6

Answer 55

thank you sososo much for your help!

Answer 56

it means a lot

Answer 57

yay team \c:/

Answer 58

hehe

Answer 59

Since youve been such a good help can you help me with one more problem?