Question

calculus

Help me solve #44 please
https://www.dropbox.com/s/r8mq9jvwwaudsut/Screenshot%202015-12-07%2014.14.19.png?dl=0

Answer 1

For first one we have x= pi and y= xcos(3x)
have u done any work so far ?

Answer 2

okay first put x=pi in y= x cos(3x)
so y = ?

Answer 3

I have solved that one, can you help me with #44 please

Answer 4

oh wait a bit ..:)

Answer 5

\[\large\rm y=\sin(1+x^3)\]So for your derivative,\[\large\rm y'=\cos(1+x^3)\cdot(1+x^3)'\]chain rule, ya?

Answer 6

yes

Answer 7

\[\large\rm y'=\cos(1+x^3)\cdot(3x^2)\]The derivative evaluated at x=-3 will give us the `slope` of our tangent line,\[\large\rm Y=mx+b\]
\[\large\rm y'(x)=3x^2 \cos(1+x^3)\]\[\large\rm y'(-3)=m\]
Plug it in.
Figure out your slope.
Do it. :)

Answer 8

whats cos(8)

Answer 9

@zepdrix

Answer 10

cos(-8)

Answer 11

(-3)^3 is not anywhere near 8 silly >.<

Answer 12

\(\rm (-3)^3\ne-9\)

Answer 13

cos(-26) ? :D

Answer 14

Ok good.
And it has a coefficient in front of it, ya?
Like a -9 or something. Or 27, ya.

Answer 15

yes

Answer 16

\[\large\rm m=27\cos(-26)\]

Answer 17

Oh we can make one nice simplification.
Since cosine is an even function, it follows this rule:\[\large\rm \cos(-x)=\cos(x)\]

Answer 18

Therefore,\[\large\rm m=27\cos(-26)=27\cos(26)\]

Answer 19

ok

Answer 20

So for our line that we're trying to construct, let's plug in what we have so far,\[\large\rm Y=\color{royalblue}{m}x+b\]\[\large\rm Y=\color{royalblue}{27\cos(26)}x+b\]

Answer 21

should we evaluate cos(26)

Answer 22

I don't know.
Are you submitting this online? Or just textbook stuff?

I think it looks a lot nicer if we leave it in exact form like this.
But if your instructions want decimal approximation we can do that.

Answer 23

its textbook stuff

Answer 24

I just wanna know how to do it

Answer 25

There is no fancy way "to do it" unfortunately :)
The angle measure of 26 radians doesn't correspond to any "special measure"
like pi/4 would for example.

So you either:
Leave it as is,
or use a calculator
^^

Answer 26

i see

Answer 27

\[\large\rm Y=27\cos(26)x+b\]To figure out this missing b value,
we need to be able to plug in a coordinate point, then we can solve for b.

Answer 28

But what point can we use? Hmm...
AH! Well recall that the tangent line "touches" the curve at this particular point!
So the tangent line Y and the curve y both share the coordinate point (-3, ? )

Answer 29

how do you know its radians and not degrees ?

Answer 30

Ah good question :)
Always assume it's radians unless you see the little degree bubble.
That's, ya

Answer 31

(-3,27*cos(26)*-3+b)

Answer 32

so pi is radians right?

Answer 33

No no.
To find our y-coordinate, we need to plug -3 back into the `original function`.

Answer 34

Yes, pi is measured in radians :3

Answer 35

(-3,sin(-26))

Answer 36

I saw people refering to pi as in 180 degrees

Answer 37

so then cos(3.14) = cos(180) in degrees

Answer 38

right?

Answer 39

Ah, yes good :)
When we're talking about the `unit circle`,
a radial distance of pi is half of the full circle, which is also 180 degrees.
So yes, that works out nicely.

Answer 40

ic

Answer 41

We'll apply the fact that sine is an odd function, so it follows this property:\[\large\rm \sin(-x)=-\sin(x)\]So that gives us our ordered pair:
\[\large\rm \left(-3,-\sin(26)\right)\]

Answer 42

\[\large\rm Y=27\cos(26)x+b\]both the curve and the tangent line share this point.
So now we'll plug this coordinate point into our tangent line, and solve for b.

Answer 43

thanks a lot for help!!

Answer 44

np