# A cup of tea has a temperature of 100 degrees Celsius. it is cooling in a room that has a temperature of 20 degrees Celsius. it cools in accordance with the following model.. T = temp in degrees...

A cup of tea has a temperature of 100 degrees Celsius. it is cooling in a room that has a temperature of 20 degrees Celsius. it cools in accordance with the following model..

T = temp in degrees Celsius

t = time in minutes

simplify the equation; `T(t)=100-70(1-(1/2)^(t/3))`

, then using the simplified equation from state the transformations based on ``

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### 1 Answer

We are given the cooling model of `T(t)=100-70(1-(1/2)^(t/3))`

We can rewrite this:

`100-70(1-(1/2)^(t/3))`

`=100-70+70(1/2)^(t/3)`

`=70(1/2)^(t/3)+30`

So the simplified model is `T(t)=70(1/2)^(t/3)+30`

The domain is `t>=0` as the physical model only makes sense for positive time.

The range is `30<T(t)<=100` . The function starts at 100, and approaches 30 as t increases without bound. Note that `70(1/2)^(t/3)>0` for all t.

** As stated, your problem probably has a typo. If the room is `20^@` C then the liquid will approach 20 degrees, not 30 degrees as t increases. **

The T intercept is where t=0 so it is 100.

There is no t-intercept as the function is always greater than 30.

There is a horizontal asymptote at T=30.

Using `(1/2)^t` as the base function the transformations are as follows:

A vertical stretch of factor 70.

A horizontal stretch of factor 1/3.

A vertical translation (shift) up 30 units.

The graph: