A certain machine depreciates in value according to the formula V (t) = 20000e^-

Question

A certain machine depreciates in value according to the formula V (t) = 20000e^-0.41 where V (t) is the value after t years. a) What happens to V (f) in the long run? b) When Is it worth 10,000 dollars? c) At what rate is the value changing with respect to time after 5 years? 2. Evaluate using substitution. integral x - 1/Squareroot x^2 - 2x + 1 dx 3. Find the area under y = xe^x from x = 0 to x = 1. 4. The population of a town is growing at a rate P' (t) = 5 + 3t^2/3 people per month and initially there are 1200 people. a) Find the formula for P (t) after t months. b) Find the increase in population over the first eight months 5. Evaluate integral^infinity _0 xe^-x^3 dx. This tells you whether the area of the region under the curve from 0 to infinity is finite or not, and if finite, tells you what it is. 6. On the surface z = x - xy + 3 find the slope in the y-direction and the slope in the x-direction at (1, 1).

Explanation / Answer

1. a> we are given that the value depreciates that is decreases according to the formula :

V(t) = 2000e^(-0.4t)

as t increases (-0.4t) increases as well.

now we can write the formula as :

V(t) = 2000/e^(0.4t)

and we know as time increases e^(0.4t) will increase because (0.4t) increases.

Hence the value V(t) will keep on decreasing as time t increases and when t = infinity the value of V(t) will become 0.

b> We need to find the time t when V(t) = 10000.

=> PLug V = 10000 in the formula and find the value of t

= V(t) = 10000 = 20000e^(-0.4t)

1/2 = e^(-0.4t)

take natural log both sides

=> ln(1/2) = -0.4t

-.693 = -.4t

=> t = 1.7325 years approximately

c> V(t) = 2000e^(-0.4t)   ----------->(1)

to find the rate differentiate (1) with respect to t

=> dV/dt = 2000e^(-0.4t)*(-0.4)    ----------->(2)

now plug t = 5 in equation (2)

=> dv/dt |t = 5 = 2000e^(-0.4*5)*(-0.4) = -108.268

hence the price is decreasing at a rate of 108.268 dollars per year

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