homework for section 4.2: linear approximations and differentials

question 9

let (y = 4x^2).
find the change in (y), (delta y) when (x = 2) and (delta x = 0.1)

find the differential (dy) when (x = 2) and (dx = 0.1)

Question

homework for section 4.2: linear approximations and differentials

question 9

let (y = 4x^2).
find the change in (y), (delta y) when (x = 2) and (delta x = 0.1)

find the differential (dy) when (x = 2) and (dx = 0.1)

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Differentials" ], "new_concepts": [ "Actual Change Calculation" ], "current_concepts": [ "Differentials", "Actual Change Calculation" ] } ## Calculate the actual change $\Delta y$ $$ \begin{aligned} y(x) &= 4x^2 \ \Delta y &= y(x + \Delta x) - y(x) \ \Delta y &= 4(2 + 0.1)^2 - 4(2)^2 \ \Delta y &= 4(2.1)^2 - 4(4) \ \Delta y &= 4(4.41) - 16 \ \Delta y &= 17.64 - 16 = 1.64 \end{aligned} $$ ## Calculate the differential $dy$ $$ \begin{aligned} dy &= y'(x) \, dx \ y'(x) &= 8x \ dy &= 8(2)(0.1) \ dy &= 16(0.1) = 1.6 \end{aligned} $$ ## Question 1 Find the change in $y$, $\Delta y$ when $x = 2$ and $\Delta x = 0.1$ 1.64 ## Question 2 Find the differential $dy$ when $x = 2$ and $dx = 0.1$ 1.6 { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Calculus", "Differentials" ] }

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Question

Explanation:

Calculate the actual change \(\Delta y\)

Calculate the differential \(dy\)

Answer:

Question 1

Question 2