Answer to Question #108087 in Algebra for rana

Question

The thickness of the ice on a lake for one week is modeled by the function:

T(d)=−0.1d3+1.2d2−4.4d+14.8

where T is the thickness in cm and d is the number of days after December 31st.

a)

Graph this function using graphing technology(desmos). Ensure your axes are properly set up, and state the domain and range that are appropriate for the situation. When do you think the warmest day occurred during the week? Justify your answer, then determine the average rate of change on a short interval near the point you chose.



b)

Determine the instantaneous rate of change at this point.



c)

Are the rates you calculated in a) and b) the same or different? Why do you think this is?

Answer 1

From the above graph we can see that value of this function becomes zero at d=8.929

hence domain of this function is [0,8.929]

Warmest day is the day when the thickness of the ice becomes zero its value is d=8.929

B)

average rate $=-\frac{\Delta T}{\Delta d}=-\frac{14.8}{8.929}=-1.6575$ the minus sign shows that thickness is decresing

instantaneous rate =$T'(d)_{8.929}=(-0.1\times3d^2+1.2\times2d-4.4)_{8.929}=(-0.1\times3(8.929)^2+2.4\times8.929-4.4)=-6.888$

C)

the average rate and instantaneous rate is different which is because instantaneous rate is it the rate of change about a particular point while average rate is rate of change over a particular period or domain