The value of a machine DECREASES at a rate proportional to the value of the machine, thus dVdt=−kVdVdt=-kV. The general solution to the differential equation is V=C⋅e−k⋅tV=C⋅e-k⋅t, where C is the initial cost of the machine.If the value of the machine is 6% of the origianl cost after 25 years,A) What is the value of k (enter as a positive value) in the differential equation? B)How many years after it is purchased will the value be half of the origianl cost?
The value of a machine DECREASES at a rate proportional to the value of the machine, thus dVdt=−kVdVdt=-kV. The general solution to the differential equation is V=C⋅e−k⋅tV=C⋅e-k⋅t, where C is the initial cost of the machine.If the value of the machine is 6% of the origianl cost after 25 years,A) What is the value of k (enter as a positive value) in the differential equation? B)How many years after it is purchased will the value be half of the origianl cost?
The value of a machine DECREASES at a rate proportional to the value of the machine, thus dVdt=−kVdVdt=-kV. The general solution to the differential equation is V=C⋅e−k⋅tV=C⋅e-k⋅t, where C is the initial cost of the machine.If the value of the machine is 6% of the origianl cost after 25 years,A) What is the value of k (enter as a positive value) in the differential equation? B)How many years after it is purchased will the value be half of the origianl cost?
The value of a machine DECREASES at a rate proportional to the value of the machine, thus dVdt=−kVdVdt=-kV. The general solution to the differential equation is V=C⋅e−k⋅tV=C⋅e-k⋅t, where C is the initial cost of the machine.
If the value of the machine is 6% of the origianl cost after 25 years,
A) What is the value of k (enter as a positive value) in the differential equation?
B)How many years after it is purchased will the value be half of the origianl cost?
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.