Question

- Differentiation Question

Answer 1

An ellipsis is defined by \[\frac{ x^2 }{ 4 } + \frac{ y^2 }{ 4 }\]

(i) Find the slopes of the tangents to this ellipse at x = 2.
(ii) For which points (x, y) are the tangents to this ellipse parallel to
the y-axis?

Answer 2

i thought loser had this one wrapped up

Answer 3

he left :'(

Answer 4

This is implicit differentiation no?

Answer 5

implicit yes :)

Answer 6

Heres my problem, if the equation given is x^3 + y^3 - 3xy = 0 for example, I can do everything. When given fractions I'm lost on how to start..

Answer 7

your equation is a circle by the way

Answer 8

a fraction is just "some number", a constant

Answer 9

My initial reaction is to find dy/dx

Answer 10

Arriving at a scenario where the y = y[x]^2

Answer 11

subbing in the given point.

Answer 12

\[g=\frac{ x^2 }{ 4 } + \frac{ y^2 }{ 4 }\]
\[g'=\frac{ 2x }{ 4 } + \frac{ 2y }{ 4 }y'\]
let x=2

Answer 13

ah your kidding me, that easy?

Answer 14

I thought you brought the 4 above the line etc

Answer 15

in your "ellipse"; if x=2, y = 0

Answer 16

if you wanted to do a quotient rule for a constant .... it wouldnt hurt nothing

Answer 17

\[\frac{x}{2}\to\frac{2-0(x)}{2^2}=\frac{1}{2}\]

Answer 18

Ah, ok.

Thats fine then, and for the 2nd part?

Answer 19

assuming that you left out something to start with:
\[\frac{x^2}{4}+\frac{y^2}{4}=1~:~ellipse\]
\[x^2 + y^2 = 4~:~circle\]

Answer 20

either way, when x=2, y=0

Answer 21

Whoops sorry, the equation is equal to 2

Answer 22

\[g'(x,y)=\frac{ 2x }{ 4 } + \frac{ 2y }{ 4 }y'\]
\[g'(2,0)=\frac{ 2(2) }{ 4 } + \frac{ 2(0) }{ 4 }y'=1\]

Answer 23

ifn =2, then thats really dividing by 8

1/2 + y^2/8 = 1

4 + y^2 = 8; y=+- 2

Answer 24

im leaving alot of details out, im assuming this makes sense :)

Answer 25

It does, do you mind if i throw a hard one at you? test that great brain of yours :P

Answer 26

I've got a big exam tomorrow, and these last couple are my only worries..

Answer 27

sites doing the lag thing again eh

Answer 28

Yeah :( Did you get the picture?

Answer 29

got the pic

Answer 30

first one, need a formula for sureface area of a sphere

Answer 31

I'm studying to be a biomedical engineer so as you imagine, diff/int are huge! Big exam tomorrow :)

Answer 32

and volume of a sphere

Answer 33

A = 4pir^2
V = 4/3pier^3

Answer 34

give me A' and V' wrt.r

Answer 35

A' = 8pir

Answer 36

V' = 4/3pi(3r^2)

Answer 37

mighta misspoke; leave the "wrt" undetermined at the moment

so both of them pop out an r' as well

Answer 38

A' = 8pir r'

V' = 4/3pi(3r^2) r'

Answer 39

we know that A' = 16, thats given

we know r = 12

that means we have all we need for V'

Answer 40

16 = 8pi (12) r'

V' = 4/3pi(3(12)^2) r'

Answer 41

\[\frac{ dA }{ dt } = \frac{ dA }{ dr } . \frac{ dr }{ dt }\] ?

Answer 42

thats one way to sprain your wrist on it yes :)

Answer 43

i get 96 if my head math is good

Answer 44

so once dA/dt was found, i can then go and say dV/dt = dv/dr x dr/dt?

Answer 45

yes

Answer 46

solve dv/dt at r = 12 ?

Answer 47

yes

Answer 48

and solve r' for r=12 and A' = 16

Answer 49

it we implicit for r(t)
\[A = 4pi~r^2\]
\[A' = 8pi~ r~r'\to r'=\frac{A'}{8pi~r}\]
\[V=\frac43pi~r^3\]
\[V'=4pi~r^2~r'\]
\[V'=4pi~r^2~\frac{A'}{8pi~r}\]
\[V'=~r~\frac{A'}{2}\]

Answer 50

therefore: V' = 12(8) = 96

Answer 51

for the next one we need volume of a cylindar
\[V = pi~r^2h\]
\[V = 2pi~rr'~h+pi~r^2h~'\]
we are given r' = 2, h'=-5
r =8, h =12

\[V = 2pi~8(2)(12)+pi~(8)^2(-5)\]

Answer 52

forgot me tick on the V' :)

Answer 53

might need to convert mm and cm to macth tho

Answer 54

1 cm = 100 mm

so for smaller values we can convert r' = .02, and h' = -.05