Question

at the given point fine the slope of the curve the line that is tangent to the curve or the line that is normal to the curve

(2x^2)y - picosy = 3 pi, tangent at (1,pi)

Answer 1

So we want to find the slope of the line tangent to the curve at (1,pi)? :D

Do you know how to differentiate implicitly? It will require us to differentiate this without solving for y explicitly beforehand.

Answer 2

I know how to differentiate implicty but not the picosy=3pi part...... What does 3pi turn in to and what is the drivitive of picos(y)

Answer 3

\[\large (\pi \cos y )'=\pi(\cos y)'= \pi(-\sin y)(y')\]
Pi is just a constant factor on the cosine, don't let it confuse you. We can ignore it while differentiating :D

Answer 4

And then on the right side, what happens when you differentiate a constant? :D Memberrrr?

Answer 5

On the right side it is just 3 right?

Answer 6

\[\huge 3 \pi \approx 9\]
What is the derivative of 9?

Answer 7

Don't let the Pi symbol confuse you, it's all just one giant constant.

Answer 8

Derivative of a constant is 0 right? :O

Answer 9

Here are a couple examples to maybe help you understand :)
These are just FANCY constants, they have their tuxedo on, trying to confuse everyone.. but their just frauds! :O Just constant! No variation!
\[\large (2 \ln 5)'=0\]\[\large (4e^2)'=0\]

Answer 10

Bu isn't the drivitive of 3pi since pi is only a constant

Answer 11

3pi is a constant, so it's derivative is 0 :O confused about that?

Answer 12

But isn't the drivitive of 3pi=3 since pi is just a contact (variable)?

Answer 13

hmm

Answer 14

pi is not a variable though :D it's just a number. it's 3.141592653689793238.....

Answer 15

Okay I get it now but when I solve the equation to the very last step I got a different answer than my textbook

Answer 16

Ok I'll go through some of the steps and you can tell me where it went wrong c:

Answer 17

Taking the derivative of both sides gives us...
Applying the product rule on the first term...
\[\large (2x^2)'y+(2x^2)y'-\pi(\cos y)'=(3 \pi)'\]

Answer 18

\[\large (4x)y+(2x^2)y'-\pi(-\sin y)y'=0\]

Answer 19

\[\large 4xy+2x^2y'+\pi (\sin y) y'=0\]

Answer 20

Factoring out a y' from each of the second terms, gives us...
\[\large y'(2x^2+ \sin y) + 4xy=0\]

Answer 21

Get confused on the product rule part or no? :O

Answer 22

= - 2pi

Answer 23

Thank you very much I get ho

Answer 24

ok cool c:

Answer 25

I get the answer now

Answer 26

I missed a pi when i factored out the y' terms but it doesn't look like it'll affect your answer c: k cool