Question

The number of barrels of oil produced from North Dakota oil wells since January 2014 is estimated to be B = 29.05e⁰.⁰²¹⁴ᵗ million barrels, where is in months since January 2014.
Estimate the relative rate of change of oil production in March 2015 using the following values of Δt. Round your answers to three decimal places.
(a) Δt = 1
(b) Δt = 0.1
(c) Δt = 0.01

Answer 1

Answer:

a) $B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t = 1$ we have the following value

$B(t+ \Delta t) = B(15+1) = B(16) = 29.05 e^{0.0214 *16}= 40.912$

And the relative change would be given by:

$\frac{B(16) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.912-40.046}{40.046}= 0.0216 = \approx 2.16 \%$

b) $B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t =0.1$ we have the following value

$B(t+ \Delta t) = B(15+0.1) = B(15.1) = 29.05 e^{0.0214 *15.1}=40.131$

And the relative change would be given by:

$\frac{B(15.1) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.131-40.046}{40.046}= 0.00214 = \approx 0.214 \%$

c) $B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t =0.1$ we have the following value

$B(t+ \Delta t) = B(15+0.01) = B(15.01) = 29.05 e^{0.0214 *15.1}=40.054$

And the relative change would be given by:

$\frac{B(15.01) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.054-40.046}{40.046}= 0.000214 = \approx 0.0214 \%$

Step-by-step explanation:

For this case we have the following function given:

$B(t) = 29.05 e^{0.0214 t}$

Where t represent the number of months since January of 2014

Part a

For this case we have the initial time at January 2014 and that represent  the value of t =0.

For march of 2015 the value fo t would be t = 15

We can estimate the corresponding value of B(15) and we got:

$B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t = 1$ we have the following value

$B(t+ \Delta t) = B(15+1) = B(16) = 29.05 e^{0.0214 *16}= 40.912$

And the relative change would be given by:

$\frac{B(16) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.912-40.046}{40.046}= 0.0216 = \approx 2.16 \%$

Part b

For this case we have the initial time at January 2014 and that represent  the value of t =0.

For march of 2015 the value fo t would be t = 15

We can estimate the corresponding value of B(15) and we got:

$B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t =0.1$ we have the following value

$B(t+ \Delta t) = B(15+0.1) = B(15.1) = 29.05 e^{0.0214 *15.1}=40.131$

And the relative change would be given by:

$\frac{B(15.1) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.131-40.046}{40.046}= 0.00214 = \approx 0.214 \%$

Part c

For this case we have the initial time at January 2014 and that represent  the value of t =0.

For march of 2015 the value fo t would be t = 15

We can estimate the corresponding value of B(15) and we got:

$B(15)= 29.05 e^{0.0214 *15}= 40.046$

For $\Delta t =0.1$ we have the following value

$B(t+ \Delta t) = B(15+0.01) = B(15.01) = 29.05 e^{0.0214 *15.1}=40.054$

And the relative change would be given by:

$\frac{B(15.01) - B(15)}{B(15)}$

And replacing we got:

$\frac{40.054-40.046}{40.046}= 0.000214 = \approx 0.0214 \%$

Answer 2

The relative rate of change of oil production in March 2015 using the given values of Δt are;

A) f'(t)/f(t) = 2.163%

A) f'(t)/f(t) = 2.163%B) f'(t)/f(t) = 0.214%

A) f'(t)/f(t) = 2.163%B) f'(t)/f(t) = 0.214%C) f'(t)/f(t) = 0.022 %

We are given the function for the estimated number of barrels of oil produced from North Dakota as;

B = 29.05e^(0.0214t)

If t is the number of months since January 2014, therefore by March 2015, t = 15 months

A) At rate of change; Δt = 1;

Formula for relative rate of change is;

f'(t)/f(t) = (1/f(t))[(f(t + Δt) - f(t)]/Δt

t + Δt = 15 + 1 = 16

Thus;

f(t + Δt) = 29.05e^(0.0214 × 16)

f(t + Δt) = 40.9118

f(t) = 29.05e^(0.0214 × 15)

f(t) = 40.0456

Plugging in the relevant values;

f'(t)/f(t) = (1/40.0456) × (40.9118 - 40.0456)/1

f'(t)/f(t) = 0.02163

f'(t)/f(t) = 2.163%

B) At rate of change; Δt = 0.1;

t + Δt = 15 + 0.1 = 15.1

Thus;

f(t + Δt) = 29.05e^(0.0214 × 15.1)

f(t + Δt) = 40.1314

f(t) = 29.05e^(0.0214 × 15)

f(t) = 40.0456

Plugging in the relevant values;

f'(t)/f(t) = (1/40.0456) × (40.1314 - 40.0456)/1

f'(t)/f(t) = 0.00214

f'(t)/f(t) = 0.214%

C) At rate of change; Δt = 0.01;

t + Δt = 15 + 0.01 = 15.01

Thus;

f(t + Δt) = 29.05e^(0.0214 × 15.01)

f(t + Δt) = 40.0542

f(t) = 29.05e^(0.0214 × 15)

f(t) = 40.0456

Plugging in the relevant values;

f'(t)/f(t) = (1/40.0456) × (40.0542 - 40.0456)/1

f'(t)/f(t) = 0.000215

f'(t)/f(t) = 0.022 %

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