Question

find the slope of the curve, the line that is tangent to the curve, or the line normal to the curve.

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

Answer 1

so would you do implicit differentiation first?

Answer 2

yeah for sure

Answer 3

would it be 4xy(y1)

Answer 4

\[ 2x^2y -\pi(\cos(y))=3\pi\]right?

Answer 5

yup

Answer 6

do you know how to do implicit diff?

Answer 7

i think so... wouldn't it be 4xy(y1) +pi(siny)(y1) ?

Answer 8

oh no
you need the product rule and the chain rule

Answer 9

think of \[2x^2y -\pi(\cos(y))=3\pi\]as if it was \[2x^2f(x) -\pi(\cos(f(x)))=3\pi\]

Answer 10

for the first term you need the product rule
\[4xf(x)+2x^2f'(x)\]

Answer 11

for the second one you need the chain rule \[-\pi \cos(f(x))f'(x)\]

Answer 12

written with \(y\) instead of \(f(x)\) it is
\[4xy+2x^2y'+\pi \sin(y)y'=0\]

Answer 13

ohhhhhh

Answer 14

so now i just plug in (1,pi)?

Answer 15

i get the implicit part now! duh. I was so confused O.o

Answer 16

you can plug in \((1,\pi)\) now and then solve for \(y'\) or you can use algebra to solve for \(y'\) first then plug in the numbers
probably easier to do the first method

Answer 17

good!

Answer 18

ok so 4(1)(pi) + 2(1)^2(y1) + (pi)sin(1)(y1)=0
4(pi) +4(y1) +

Answer 19

oh jeez i'm stuck so (pi)sin(1)(y1) would be (pi)(pi/2)(y1)?

Answer 20

\[4xy+2x^2y'+\pi \sin(y)y'=0\] at \((1,\pi)\)
\[4\pi+2y'+\pi\sin(\pi)y'=0\]

Answer 21

since fortunately \(\sin(\pi)=0\) this is just
\[4\pi+2y'=0\] and that is easy to solve for \(y'\)

Answer 22

ohh i messed up and did sin(1) whoops

Answer 23

yeah you don't know that one for sure

Answer 24

y1 =-2pi?

Answer 25

that is what i get, yes

Answer 26

yaaas !! thanks so much

Answer 27

yw

Answer 28

urmm are u busy cuz i have 1 more quick question?

Answer 29

sure go ahead and ask

Answer 30

ok this question is "If the fuction is not differentiable at the given value of x, tell whether the problem is a corner, cusp, vertical tanget or discontinuity"

y= -7 -\[\sqrt[3]{x}\] at x=0

so wouldn't I just take the derivitive which is 1/3(x^3) ? and plug in 0 ?

Answer 31

it's y= -7 - 3sqrtx

Answer 32

your derivative is wrong

Answer 33

*slams head on desk*

Answer 34

\[y=7-\sqrt[3]{x}\]
\[y'=-\frac{1}{3\sqrt[3]{x^2}}\]