Question

i dont understand this problem. HELP!
find d^2y/dx^2' if x=1+t^2 y= t-t^3

Answer 1

x=1+t^2

what is dx/dt ?

Answer 2

is it 2t?

Answer 3

yes

Answer 4

how about dy/dt ?

Answer 5

-3t^2

Answer 6

so all its asking for is the derivatives for both x and y?

Answer 7

close, but not quite there on dy/dt

Answer 8

oh 1-3t^2

Answer 9

so the FIRST derivative of y (with respect to x) is

\[\Large \frac{dy}{dx} = \frac{dy/dt}{dx/dt}\]
\[\Large \frac{dy}{dx} = \frac{1 - 3t^2}{2t}\]

Answer 10

but we want to find \[\Large \frac{d^2y}{dx^2}\] which is the second derivative of y with respect to x

Answer 11

oh second derivatives

Answer 12

would it be dy/dx = -6t/2?

Answer 13

how are you getting -6t/2 ?

Answer 14

the 3 times 2 from 3t^2?

Answer 15

I see, unfortunately it's incorrect

Answer 16

oh no ;(

Answer 17

I'm going to use the ideas given on this page

http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

so read it over. I'm going to focus on the section "Second Derivative for Parametric Equations"

Tell me when you have read it over so I know when to continue

Answer 18

i have read over it @jim_thompson5910

Answer 19

alright

Answer 20

so do you see in the blue box where it says "Second Derivative for Parametric Equations"

Answer 21

thats a good website is it yours?

Answer 22

and do you understand how that formula works?

and no it's not mine

Answer 23

yes but why is the two negative?

Answer 24

two negative?

Answer 25

where are you at?

Answer 26

example 3

Answer 27

under the blue box

Answer 28

He skipped a bunch of steps, but he used the quotient rule

Answer 29

oh

Answer 30

so do you always have to factor out the bottom part

Answer 31

you mean how he has the \(\Large (5t^3-12t)^2\) ?

Answer 32

yes

Answer 33

that's just a result of the quotient rule

Answer 34

anyways, if

\[\Large y = \frac{1 - 3t^2}{2t}\]

then what is dy/dt ?

Answer 35

is it (1-3t^2) * 2t^-1

Answer 36

yeah you can convert to that form, and then derive that with respect to t (use product rule)

Answer 37

be careful though, I think you meant to say (2t)^(-1) instead of just 2t^(-1)

Answer 38

yeah and i got -3

Answer 39

\[\Large y = \frac{1 - 3t^2}{2t}\]
\[\Large y = (1 - 3t^2)(2t)^{-1}\]
\[\Large \frac{dy}{dt} = ???\]

Answer 40

-3 is not correct

Answer 41

that works, now derive that

Answer 42

is it right?

Answer 43

you derived the 2t^(-1) incorrectly

Answer 44

it should be -2t^(-2)