Question

At time t, in hours, a lake is covered with ice of thickness y cm, where y=0.5 t^(1.4). How fast is the ice forming when t=1? t=2? answer in units

Answer 1

plug the values of t into the equation for t and solve.

Answer 2

ok so my answers would be
t=1 --> 1/2
t=2 --> 1.3195
??

Answer 3

You want to know, how fast is ice forming at t = 1, so you would need the derivative, which gives us the rate of change, at t=1.

Answer 4

Plugging in t = 1 does not tell us how fast ice is forming at t = 1.

Answer 5

Yes, plugging in t=1 does not tell you how fast ice is forming, it tells you how thick the ice is after "t" hours. To get the rate of ice formation, you have to take the derivative first, THEN plug in.

Answer 6

dy/dt = (0.7)(t^(0.4))...that is the rate of change at time t.

when t = 1, dy/dt = 7/10

Answer 7

when t = 2, dy/dt = .924

Answer 8

units are cm/hr

Answer 9

thank you @Easyaspi314 and @jb1515g

Answer 10

welcome.

Answer 11

How could I determine when ice is the thickes and when it is forming fastest from interval 0<t<3??

Answer 12

The language of "how fast is something happening" means we want the rate of change.

Answer 13

When is forming the fastest? Then we want to maximize the derivative.

Answer 14

Whats important is to understand the notion of "rate of change" and that represents the derivative.

For that matter, rate of change is just the slope of a graph. Which the derivative is just that. The slope.

Answer 15

The rate of change of the line y = 6x - 10 is 6. And that is the derivative of the function.

Answer 16

Am I talking to myself? lol

Answer 17

@Easyaspi314 still here

Answer 18

anyways, I got to leave. I could keep on going on..but let's stop here.

Answer 19

@Easyaspi314 alright, thanks

Answer 20

welcome. Have a good night.