Question

In a simple physical apparatus, the distance x cm and y cm are related by the equation 1/x + 1/y = 1/5 . if y is
increasing at a rate of 16 cm per minute, calculate the rate at which x is changing when x=15cm. [Note:
apparatus: Equipment designed to serve a specific function] [Oct08: -64cm/min]

Answer 1

how to solve it?

Answer 2

hold on

Answer 3

hmm.. i got dx/dt = -4cm/min but i dont know if its correct or not

Answer 4

if its correct then tell me and ill post up the worked solution

Answer 5

it's ok coz smetimes the ans given is incorrect.

Answer 6

dx/dt = -25(x-5)^2 times 16, then sub x=15 and u get -4cm/sec

Answer 7

no its -64...hold on

Answer 8

ok dumbcow

Answer 9

ok i got it, dx/dt = (16*(x-5)^2)/-25
sub x=15 and you get -64cm/min

Answer 10

i got x'=-576 unless i made a mistake

Answer 11

thegreatsaiyaman: can u explaim to me how u get the ans?

Answer 12

\[\frac{dx}{dt} = \frac{dy}{dt}*\frac{dx}{dy}\]
if you solve the equation for x
\[x = \frac{5y}{y-5}\]
\[\frac{dx}{dy} = -\frac{25}{(y-5)^{2}}\]
\[\frac{dx}{dt} = -\frac{16*25}{(y-5)^{2}}\]
Now they don't tell us what y is, only that x=15
so substitute 15 in to find the corresponding y-value
\[\frac{1}{y} = \frac{1}{5}-\frac{1}{15} = \frac{2}{15} \rightarrow y = \frac{15}{2}\]
\[\frac{dx}{dt} = -\frac{16*25}{(7.5-5)^{2}} = -64\]

Answer 13

*can u explain

Answer 14

dumbcows got it, i just dont know how to use these symbols on this website =.=

Answer 15

omg darn it i made a mistake

Answer 16

i forgot to write the 1 in front of 5 lol

Answer 17

huhu.. thanks all~

Answer 18

15/2 not 5/2 is y

Answer 19

yes i get x'=-64 as well lol

Answer 20

<3 calculas

Answer 21

hey bell if you want to see the way i did it here you go

Answer 22

how 5y become 25?

Answer 23

from taking the derivative.
using the quotient rule
\[\frac{f'g - fg'}{g^{2}}\]

Answer 24

ouh... ok..

Answer 25

thanks again to all~