(c) Equating the values of the car and Steven's savings: To find the time when Steven can buy the car, we need to solve the equation: V(t) = S(t) Substituting the expressions we found: 55,000 This equation can't be solved analytically, so let's use numerical methods to approximate the value of t One common numerical method is the Newton-Raphson method. We start by rearranging the equation to have zero on one side: 55,000 - 120€ - 9,000=0 Now, let's define a function f(t) = 55,000-120t -9,000. We want to find the value of t where fit) is equal to zero. 0.01L We can start with an initial guess fort, such as t=1, and then use the Newton-Raphson method to iteratively refine our approximation. The Newton-Raphson iteration formula is given by: [(L₂) P(1₂) t+=t₂ Where t, is the current approximation, f(ta) is the value of the function at to and f'(t.) is the derivative of the function at t = 120€ +9,000 Let's calculate the derivative of fit) first. f'(t)=-550e--120 t₁ = to = 1 Now, we can start the iteration process: Initial guess: t, =1 t₂ = t₁- [(4) P(4) 55,041,201-9,000 55.01.20 Performing the calculations: t₁2.376 We continue this process by t, back into the equation: ₂ 2.381 That [(4) P(4) We repeat these iterations until we reach a desired level of accuracy. Let's continue the calculations: t2.381 ₁2.381 We can see that the value of t stabilizes around t 2.381.
(c) Equating the values of the car and Steven's savings: To find the time when Steven can buy the car, we need to solve the equation: V(t) = S(t) Substituting the expressions we found: 55,000 This equation can't be solved analytically, so let's use numerical methods to approximate the value of t One common numerical method is the Newton-Raphson method. We start by rearranging the equation to have zero on one side: 55,000 - 120€ - 9,000=0 Now, let's define a function f(t) = 55,000-120t -9,000. We want to find the value of t where fit) is equal to zero. 0.01L We can start with an initial guess fort, such as t=1, and then use the Newton-Raphson method to iteratively refine our approximation. The Newton-Raphson iteration formula is given by: [(L₂) P(1₂) t+=t₂ Where t, is the current approximation, f(ta) is the value of the function at to and f'(t.) is the derivative of the function at t = 120€ +9,000 Let's calculate the derivative of fit) first. f'(t)=-550e--120 t₁ = to = 1 Now, we can start the iteration process: Initial guess: t, =1 t₂ = t₁- [(4) P(4) 55,041,201-9,000 55.01.20 Performing the calculations: t₁2.376 We continue this process by t, back into the equation: ₂ 2.381 That [(4) P(4) We repeat these iterations until we reach a desired level of accuracy. Let's continue the calculations: t2.381 ₁2.381 We can see that the value of t stabilizes around t 2.381.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
for part (c), my answer is different to your answer. is my solution correct or is it yours?
Please do answer this follow-up question. So I can ensure I have the right solution.
Transcribed Image Text:(c) Equating the values of the car and Steven's savings:
To find the time when Steven can buy the car, we need to solve the equation:
V(t) = S(1)
Substituting the expressions we found:
55,000
0.01L
This equation can't be solved analytically, so let's use numerical methods to approximate the value of t
One common numerical method is the Newton-Raphson method. We start by rearranging the
equation to have zero on one side:
55,000
-120€ - 9,000=0
Now, let's define a function f(t) = 55,000-120t -9,000. We want to find the value of t
where fit) is equal to zero.
We can start with an initial guess fort, such ast=1, and then use the Newton-Raphson method to
iteratively refine our approximation.
The Newton-Raphson iteration formula is given by:
[(L₂)
P(1₂)
= 120€ +9,000
t+=t₂
Where t, is the current approximation, f(ta) is the value of the function at to and f'(t.) is the
derivative of the function at t
Let's calculate the derivative of fit) first.
f'(t)=-550e--120
t₁ = to
= 1
Now, we can start the iteration process:
Initial guess: t=1
[(4)
P(4)
55,041,201-9,000
55.01.20
t₂ = t₁-
Performing the calculations:
t₁2.376
We continue this process by t, back into the equation:
[(4)
P(4)
₂ 2.381
We repeat these iterations until we reach a desired level of accuracy. Let's continue the calculations:
t2.381
₁2.381
We can see that the value of t stabilizes around t 2.381.
Therefore, the approximate time when Steven can buy the car ist & 2.381 months.
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