Question

At time t = 0, a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius (°C) at time t = 0, and the internal temperature of the potato is greater than 27°C for all times t > 0. The internal temperature of the potato at time t minutes can be modeled by the function H that satisfies the differential equation (dH/dt)=−(1/4)(H − 27), where H(t) is measured in degrees Celsius and H(0)=91.

(a) Write an equation for the line tangent to the graph of H at t = 0. Use this equation to approximate the internal temperature of the potato at time t = 3.
a) H(t) = -t/4 + 91
b) H(t) = -t/4 + 27
c) H(t) = t/4 + 91
d) H(t) = t/4 + 27

Answer 1

The equation for the line tangent to the graph of H at t = 0 is H(t) = -t/4 + 91. The internal temperature of the potato at time t = 3 is approximately 90.25 degrees Celsius.

Step-by-step explanation:

The differential equation given to model the internal temperature of the potato is: (dH/dt) = -(1/4)(H - 27), where H(t) is measured in degrees Celsius and H(0) = 91. To find the equation for the line tangent to the graph of H at t = 0, we need to find the derivative of H with respect to t. The derivative is (dH/dt) = -(1/4)(H - 27), which represents the slope of the tangent line. Since the tangent line passes through the point (0, 91), we can use the slope-intercept form of a linear equation to find the equation for the tangent line, which is H(t) = -t/4 + 91. To approximate the internal temperature of the potato at time t = 3, we can substitute t = 3 into the equation for the tangent line: H(3) = -3/4 + 91 = 90.25 degrees Celsius.