Understanding Differential Equations in Potato Temperature Modeling

The video tutorial explains how to solve a separable differential equation to model the internal temperature of a potato over time. It begins by introducing the problem and the differential equation, then demonstrates the process of solving it through separation of variables and integration. The tutorial finds the constant of integration using initial conditions and derives an expression for the temperature function G(t). Finally, it calculates the internal temperature of the potato at T=3 minutes.

What is the initial internal temperature of the potato at time t=0?

• 91°C

• 54°C

• 64°C

• 27°C

What type of differential equation is used in the model?

• Non-linear

• Separable

• Homogeneous

• Linear

What is the purpose of separating variables in a differential equation?

• To simplify the equation

• To integrate both sides separately

• To eliminate constants

• To find the derivative

What mathematical technique is used to solve the separated differential equation?

• Integration

• Differentiation

• Approximation

• Substitution

What is the value of the constant C found using the initial condition?

• 64

• 4

• 12

• 27

What is the significance of the initial condition G(0) = 91 in solving the differential equation?

• It helps to find the derivative

• It determines the type of equation

• It is used to solve for the constant C

• It provides the final solution

What is the expression for G(t) derived from the differential equation?

• G(t) = (t/3 + 4)^3 + 27

• G(t) = (t/3 - 4)^3 + 27

• G(t) = (t/3 - 4)^2 + 27

• G(t) = (t/3 + 4)^2 + 27

What is the exponent used in the expression for G(t) after integration?

• 1/2

• 1/3

• 2/3

• 3/2

What operation is performed after dividing both sides by 3 in the expression for G(t)?

• Taking the square root

• Taking the logarithm

• Taking the cube

• Taking the derivative

What is the internal temperature of the potato at time t=3?

• 54°C

• 91°C

• 27°C

• 64°C