Question

At time t, in hours, a lake is covered with ice of thickness y cm, where y=0.3t^1.5
1)How fast is the ice forming when t=1? Give units

2) If ice forms for 0_ 9 years ago

Answer 1

1. just substitute the value of 1 for t inside this formula and will get the right result

Answer 2

But what would the units be? It's not cm

Answer 3

It says in the question in hours right?

Answer 4

No it says hours aren't the correct answer either. I believe the right course of action is to find the derivative and then plug in 1 but I don't know what the units would be for it then

Answer 5

units would be cm/hr

Answer 6

and i'm talking about the 1st question
i would differentiate y=.3t^(1.5)
and then plug in 1
and your answer attach the cm/hr as the unit

Answer 7

It is also given y is in cm... so it definitely is cm/hr. And yeah that is only for the first one.

Answer 8

You're right it was cm/hr! Can you help me on the second part of the problem?

Answer 9

you are finding the maximum for y=.3x^(1.5) on the interval [0,3]

Answer 10

I'm not completely sure how to do that

Answer 11

do you know how to find any critical numbers?
if so are there any? and if so what are they?

Answer 12

To find critical values you take the derivative and equate it to 0 and then determine the x values correct? The only x value I could find to make the equation true was 0

Answer 13

cool stuff
you must determine for which values in [0,3] you have a maximum output for function g(t)=0.3t^(1.5)

you found critical number t=0 but that happened to be an endpoint of our interval in which we are trying to determine the max

so just evaluate g(0) and g(3)
tell me which gives us the highest number of those two

Answer 14

The higher is g(3)

Answer 15

so 3 hours the time in which we have the thickest ice

Answer 16

for the given interval anyways

Answer 17

Okay that makes sense, how could I tell when the ice is forming the fastest?

Answer 18

We need to maximize y' on [0,3]
that means we need to maximize the function h(x)=.45*t^(.5)

Answer 19

I'm just calling y' h(x)

Answer 20

so find h'
and determine critical numbers on [0,3]
if there are any you do h(critical numbers)
no matter if there are or aren't critical numbers you must also do h(end points) like we did in last problem

Answer 21

I keep making t x
I meant h(t)=.45*t^(.5) <---this was just the derivative of y=.3t^(1.5)

Answer 22

I found h' and then plugged in both 0 and 3 and 3hrs is the correct answer. Thank you so much!!!!

Answer 23

np