Question

@zepdrix Find the slope of the tangent line to the graph of 4x^2 y- pi cos y=5 pi at the point (1, pi).

Answer 1

\[\large 4x^2y-\pi \cos y=5\pi\]
Hmm ok this will work out similar to the normal line problem we did earlier.
We won't have to worry about the negative reciprocal nonsense this time though.

Answer 2

lol that's better

Answer 3

Have you tried taking the derivative yet?

Answer 4

No, not yet! I got a little confused cause there are so many different variables

Answer 5

For the first term, we'll apply the product rule.\[\large \left(4x^2y\right)' \qquad=\qquad (4x^2)'y+4x^2(y)'\]

Answer 6

Okay

Answer 7

So the first term when we take it's derivative is giving us,\[\large 8xy+4x^2y'\]
Uh oh, you're poppin out the "okay" 's. lol :) confused i'll bet

Answer 8

No I'm good, I'm working it on my paper as well so I understand! So far so good!

Answer 9

You're seriously so awesome at teaching me, you don't even know!

Answer 10

In the next term, that pi is just a constant.
So let's not worry too much about it.
Derivative of -cosine is umm positive sine right?\[\large \left(-\pi \cos y\right)' \qquad=\qquad \pi (\sin y)y'\]
And we need to recall that any time we take the derivative of something containing y, we plop a y' onto the end of it.
I put brackets around the sin y just so there's no confusion that the y' is NOT inside of the sine function with y.

Answer 11

So far so good!

Answer 12

The right side is constant.
Derivative of a constant? :O

Answer 13

Is 0!

Answer 14

Ok sounds good! :)
So this is what we have so far,\[\large 8xy+4x^2y'+\pi(\sin y)y'=0\]

Answer 15

I'm glad we did those other two problems BEFORE this one.
Because we'll be using techniques from both of those problems. :)
So it's a nice lead-in to this one.

Answer 16

Lol yay! :)

Answer 17

Let's get the y' terms on one side.
So let's subtract 8xy from each side,\[\large 4x^2y'+\pi(\sin y)y'=-8xy\]Then we'll factor a y' out of each term,\[\large y'\left(4x^2+\pi \sin y\right)=-8xy\]Then divide by some stuff.

Answer 18

So I would I just put in the (1, pi)?

Answer 19

\[\large y_t=mx+b\]

Our particular slope \(\large m\) is given by,\[\large y'(1,\pi)=m\]
Yes good :)

Answer 20

So the answer is about 2 pi! Wow so helpful! :D

Answer 21

haha `about` 2pi? XD

Answer 22

Yeah, I got a decimal lol, not exactly 2 pi

Answer 23

Hmm weird, I got -2pi.

Answer 24

I mean negative! But did you get that exactly?

Answer 25

\[\large y'(x,y)=\frac{-8xy}{4x^2+\pi \sin y}\]
\[\large y'(1,\pi)=\frac{-8(1)(\pi)}{4(1)^2+\cancel{\pi \sin \pi}}\]

Answer 26

\[\large y'(1,\pi)=\frac{-8\pi}{4}\]Hmm ya I got exactly :o

Answer 27

Sine of pi is zero, did you punch it into your calculator in degree mode or something silly like that?

Answer 28

Lol cause I put it in the calculator, but the answer doesn't come out as pi you have to convert it unless there is some button I don't know about. But doing it by hand like you did helps too!

Answer 29

oh i see

Answer 30

Okay, I just have a few more I promise! You are really helping me do good on my test tomorrow! I thought I would have to be up for hours trying to figure these out!

Answer 31

Ok post another one, then imma take a lil math break after that :D
Lots of people on here to help you though, don't worry!

Answer 32

No one seems to respond or know how to do them lol