Question

Rate of change problem revisited:

Answer 1

Would REALLY like some step by step for this one :) @phi

Answer 2

I already posted on this
start with
A= pi r^2

take the derivative with respect to time
dA/dt = 2 pi r dr/dt
find the numeric value of dA/dt (how fast the area is increasing) for a given r and dr/dt
multiply by 2 (for 2,000,000 rand per square km)

Answer 3

The idea in these problems is first get an equation that relates the variables.
Here you want how fast the area is changing (i.e. how fast the forest is being consumed by the fire). They tell you how fast the radius is changing. (dr/dt)

that suggests find an equation relating area and radius.
A= pi r^2

take the derivative. the derivative of A with respect to time t is just dA/dt
the derivative of r^2 with respect to t is 2 r dr/dt
they tell you r and dr/dt so you can find dA/dt

Answer 4

yeah so i have dA/dt = 2 pi r dr/dt
and dr/dt = 0.1 [as r = 2 + 0.1t]
so dA/dt = 2 pi r (0.1) = 0.2 pi r
so dR/dt = dR/dA x dA/dt = 2 x (0.2 pi r) = 0.4 pi r
which is wrong

Answer 5

where did i go wrong?

Answer 6

(the R above is Rands lost)

Answer 7

and dr/dt = 0.1 [as r = 2 + 0.1t]

I don't follow the [as r = 2 + 0.1t]
they tell you dr/dt is 0.1 km/hour

Answer 8

so it doesnt matter that it started off at 2km radius?

Answer 9

so dA/dt = 2 pi r (0.1) = 0.2 pi r

true. they are asking for dA/dt when r is 2 km
so replace r with 2 km
dA/dt =0.2 km/hour pi 2 km

Answer 10

you get dA/dt = 0.4 pi km^2/hour
now mult by 2 to get millions of rands per square km per hour

Answer 11

o...okay. so my answer of 0.4 pi r just needed a 2 to be subbed in for r?

Answer 12

thanks A LOT :)

Answer 13

much appreciated

Answer 14

all 3 of your problems follow this same pattern. Notice you don't really use the chain rule. It is more the idea of taking "implicit" differentials

if you are used to
\[ \frac{d}{dx} x^2 = 2 x\]
think more like
\[ d (x^2) = 2x dx \]

or with respect to t:
\[ \frac{d}{dt}x^2 = 2 x \frac{dx}{dt} \]