Post

Created by @nathanedwards
 at November 3rd 2023, 9:51:27 pm.

Question:

The temperature of a liquid is changing with respect to time. The formula that models the change in temperature (in degrees Celsius) over time (in minutes) is given by:

T(t)=2t35t2+8t3T(t) = 2t^3 - 5t^2 + 8t - 3

Find the linear approximation of the temperature at t=2t = 2 minutes. Use this approximation to estimate the temperature at t=2.1t = 2.1 minutes. Round your answer to two decimal places.

Answer:

To find the linear approximation of the temperature at t=2t = 2 minutes, we will use the concept of the tangent line to approximate the temperature near that point. The equation of the tangent line is given by:

L(t)=f(a)+f(a)(ta)L(t) = f(a) + f'(a) \cdot (t-a)

where f(a)f(a) represents the value of the function at t=at = a, and f(a)f'(a) represents the derivative of the function evaluated at t=at = a. In our case, a=2a = 2.

First, let's find the value of the temperature at t=2t = 2 minutes:

T(2)=2(2)35(2)2+8(2)3=1620+163=9T(2) = 2(2)^3 - 5(2)^2 + 8(2) - 3 = 16 - 20 + 16 - 3 = 9

So, f(2)=9f(2) = 9.

Now, let's find the derivative of the temperature function:

T(t)=ddt(2t35t2+8t3)T'(t) = \frac{d}{dt} (2t^3 - 5t^2 + 8t - 3)

Applying the power rule and the sum rule of derivatives, we get:

T(t)=6t210t+8T'(t) = 6t^2 - 10t + 8

Next, we need to evaluate the derivative at t=2t = 2:

T(2)=6(2)210(2)+8=2420+8=12T'(2) = 6(2)^2 - 10(2) + 8 = 24 - 20 + 8 = 12

So, f(2)=12f'(2) = 12.

Now, let's substitute these values into the equation of the tangent line to find the linear approximation at t=2t = 2 minutes:

L(t)=f(2)+f(2)(t2)L(t) = f(2) + f'(2) \cdot (t-2)
L(t)=9+12(t2)L(t) = 9 + 12 \cdot (t-2)

Simplifying the expression:

L(t)=3+12tL(t) = -3 + 12t

Therefore, the linear approximation of the temperature at t=2t = 2 minutes is L(t)=3+12tL(t) = -3 + 12t.

To estimate the temperature at t=2.1t = 2.1 minutes using this linear approximation, we substitute t=2.1t = 2.1 into the equation of the tangent line:

L(2.1)=3+12(2.1)L(2.1) = -3 + 12(2.1)
L(2.1)=3+25.2L(2.1) = -3 + 25.2
L(2.1)=22.2L(2.1) = 22.2

Therefore, the estimated temperature at t=2.1t = 2.1 minutes is approximately 22.2 degrees Celsius.