Question

If f and g are functions of time, and at time t = 2, f equals 5 and is rising at a rate of 6 units per second, and g equals 7 and is rising at a rate of 9 units per second, then f/g equals and is changing at a rate of units per second. If f and g are functions of time, and at time t = 3, f equals 9 and is rising at a rate of 6 units per second, and g equals 8 and is rising at a rate of 9 units per second, then f/g equals and is changing at a rate of units per second.

Answer 1

Answer:

f/g at t =3 will be 9/8

d/ddx(f(x)/g(x))=f'(x)d(x)-f(x)g'(x)/(g'(x))^2

rate of change=(6*8)+(9*9)/9^2

rate of change =43/27=1.5925

Answer 2

Answer:

a) $\frac{f}{g}= \frac{5}{7}$

It is changing at a rate of $\ \frac{-3}{49}$  units per seconds

b) $\frac{f}{g}= \frac{9}{8}$

It is changing at a rate of $\frac{- \ 33}{64}$ units per seconds

Step-by-step explanation:

A

Given that:

f and g are functions of time (t):

i.e  

f(t) =0

g(t) = 0

Now; at time (t) = 2 ; f equals 5 and is rising at a rate of 6 units per second.

Also; at time 2; g equals 7 and is rising at a rate of 9 units per second

This implies that:

f(2) = 5                   f' (2) = 6

g(2) = 7                  g' (2) = 9

Then $\frac{f}{g}$ can be directly  calculated from the  given values which is:

$\frac{f}{g}= \frac{5}{7}$

Using differentiation to determine the rate of changing ; we have

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=2}$ $= \frac{g(t).f'(t)-f(t).g'(t)}{(g(t))^2}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=2}$ $= \frac{7.(6)-5.(9)}{(7)^2}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=2}$ = $\frac{42-45}{49}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=2}$ $= \ \ \ \frac{-3}{49}$

∴ It is changing at a rate of $\ \frac{-3}{49}$  units per seconds

B

Given that:

f and g are functions of time (t):

i.e  

f(t) =0

g(t) = 0

NOW; at time (t) = 3

f (3) = 9                                      f' (3) = 6

g (3) = 8                                     g' (3) = 9

Then $\frac{f}{g}$ can be directly  calculated from the  given values which is:

$\frac{f}{g}= \frac{9}{8}$

Using differentiation to determine the rate of changing ; we have

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=3}$ $= \frac{g(t).f'(t)-f(t).g'(t)}{(g(t))^2}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=3}$ $= \frac{8.(6)-9.(9)}{(8)^2}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=3}$ $= \frac{48-81}{64}$

$\frac{d}{dt} [\frac{f(t)}{g(t)} ]_{t=3}$ $= \ \frac{- \ 33}{64}$

∴ It is changing at a rate of $\frac{- \ 33}{64}$ units per seconds

The negative signs means they are decreasing.