applications of calculus.. (1 Viewer)

[sukiyaki]

(rates) , i know the question it easy but any question with no equation makes me completely
lost where to start

+ a cubical block of ice has an edge of 10cm. It melts so that its volume decreases at a constant
rate and the block remains cubical. If it edge measures 5cm after 70 minutes find
(A) the rate at which the volume decreases
(B) the volume at any time t

(from fitzpatrick)

thanks in advance =)

 

[wogboy]

a)

volume of a cube= (side length)^3

so,

initial volume = 10^3 cm^3 = 1000 cm^3
final volume = 5^3 cm^3 = 125 cm^3

given that the volume decreases at a constant rate,

rate of change of volume
= (final volume - initial volume)/(time taken)
= (125 - 1000)/70 cm^3/min
= -12.5 cm^3/min (this is negative since the volume is decreasing)

so the rate at which the volume decreases is = 12.5 cm^3/min

(positive, since the rate of decrease is the negative of the rate of change. Rate of change implies that the quantity is increasing)

b)

to find the volume at time t, we'll assume t=0 when the cube is full size (i.e. has side length of 10 cm)

therefore,

volume = initial volume + t *(rate of change of volume)
= 1000 + t*(-12.5)

therefore,

volume = 1000 - (12.5)t cm^3 (where t is in minutes)

(of couse when t > 8 min, the cube totally melts and nothing is left of it, i.e. volume is zero, you can't have negative volume).

Strange question, doesn't require the use of calculus at all!

Tip: Whenever you see questions like this, the first thing to do is to draw a diagram. Then you'll understand the question, and hopefully the formulas (formulae?) will come to you.

 

[sukiyaki]

oO!! thanks wogboy =)

 

[TimTheTutor]

Originally posted by wogboy


Strange question, doesn't require the use of calculus at all!

Here is a method that does use calculus (integration & constant)

(a)
dV/dt = K (K<0 since the volume constantly decreases over time)
Integrating with respect to t
V = Kt + C
x^3 = Kt + C
When t=0 x=10 => C=1000
x^3 = 1000 + Kt
When t=70 x=5
125 = 1000 + 70K
70K = -875
K = -12.5 m/s

(b) V = -12.5t + 1000 (0<=t<=80)