Question

Parametric Differentiation

x= sqrt{t} , y = sqrt{t-1}

t = 2

Find dy/dx and evaluate the specified value of the parameter

Answer 1

alright what throws me off is idk if to just make the y equation the sqrt of t and differentiate it using the constant rule or to say (t-1)1/2 then differentiate we're only concerned with t so its just the sqrt of t right for my y equation?

Answer 2

derive both parts seperately and then put the y over the x and solve with t

Answer 3

x = sqrt{t} y = sqrt{t-1}

dx = 1/2sqrt(2) dy = 1/2sqrt{t-1}

Answer 4

sqrt(2) means sqrt(t) :)

Answer 5

dont we get rid of the 1 or kinda ignore it?

Answer 6

no, its a part of the function

Answer 7

the innards derive to simply '1' so it doesnt affect the normal flow of the deriving

Answer 8

so it'll be1/2 (t-1)-1/2(1)

Answer 9

now if it had been:

y = sqrt(2x+1) ; then that derives to:

D(2x+2)
----------
2sqrt(2x+1)

Answer 10

\[{dy \over dx}=\frac{1/2\sqrt{t-1}}{1/2\sqrt{t}}\]

Answer 11

i know how to do the bottom im not sure about the top

Answer 12

because my lecturer said were not concerned about the values with no t

Answer 13

but that one is different

Answer 14

get this looking normal and then rationalize the denom

Answer 15

\[\frac{2\sqrt{t}}{2\sqrt{t-1}}\]

Answer 16

\[\frac{\sqrt{t}}{\sqrt{t-1}}*\frac{\sqrt{t-1}}{\sqrt{t-1}}\]

Answer 17

it might just have to end up having a negative square root .... :)

Answer 18

why'd you put the two in fornt of the sqrt{t}

Answer 19

\[\frac{\sqrt{t^2 -t}}{t-1}\]

Answer 20

because sqrt(t) derives to 1/2sqrt(t)

Answer 21

now when t=2 we get: sqrt(2)/2

Answer 22

er..... \[\sqrt2\]

Answer 23

look over what I got, I gotta go to class for a few hours :) good luck

Answer 24

ok thank you so much! have fun!!!