Question

what is the second derivative of the parametic function x = sqrt(t), y =8t-1, t= 4, if the first derivative is equal to 32? ( I did the math for the first deriv., but I cannot get the second one.)

Answer 1

Hello!

Answer 2

did you get first derivative:
\[=16\sqrt{t}\]
?

Answer 3

No, I got the derivative as 8/(1/2t^1/2)

Answer 4

or 8/(1/4)

Answer 5

I got for second derivative:
\[d/dt (dy/dx)/(dx/dt)\]
\[d(16\sqrt{t})dt/(dx/dt)=16\]

Answer 6

But i don't think I did it right, even if 32 is the right answer

Answer 7

i just went with definition of second derivative and fact that
y=8t-1=8x^2-1 (put the value of x=sqrt t)

Answer 8

I had 1/2t^(-1/2) as the dx/dt for the first derivative

Answer 9

I got 16.... so I missed *2 somewhere... let me check

Answer 10

and y = 8 for dy/dt

Answer 11

\[dy/dx=(dy/dt)/*dx/dt)=8/(1/2\sqrt{t})=16\sqrt{t}\]
right?

Answer 12

...

Answer 13

yes, but I

Answer 14

i missed typed denominator... shouldn't be any "*"

Answer 15

wouldn't it be 16/ sqrt(t)?

Answer 16

\[(\sqrt{t})'=1/(2\sqrt{t})\]

Answer 17

?

Answer 18

But even with 16/sqrt(t), you would but in 4 and get 16/2 right? Which is 8 though

Answer 19

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Answer 20

let's try one more time:
\[dy/dx=(dy/dt) / (dx/dt)=8/(1/2\sqrt{t})=16\sqrt{t}\]

Answer 21

you mean 16/sqrt(t)

Answer 22

no. there is no division.

Answer 23

????

Answer 24

okay, so the 1/2 multiplied to sqrt(t) doesn't get moved to the top

Answer 25

\[8/(1/(2*\sqrt{t}))=8*2*\sqrt{t}\]

Answer 26

so, if 16sqrt(t), then 16* sqrt(4) = 16*2 is 32. cool. we got it. Now, the second derivative though?

Answer 27

second derivative\[=d/dt (dy/dx) : dx/dt\]
\[y=8t-1=8x ^{2}-1\]
(use substitution from first eq)
agreed?

Answer 28

where did the x come from?

Answer 29

\[x=\sqrt{t}\]
given

Answer 30

okay, yep agreed

Answer 31

so, 16x?

Answer 32

yes it is!!!! thanks for all your help!:D

Answer 33

awesome!