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Linearization & Differentials - Interactive Lesson | Study Mondo
Linearization & Differentials - Complete Interactive Lesson Part 1: The Tangent Line Approximation Linearization & Differentials
Part 1 of 7 โ The Tangent Line Approximation
Topic Overview
Part Topic 1 Tangent line approximation 2 Approximating values 3 Differentials 4 Error analysis 5 Applications & related rates 6 AP-style workshop 7 Comprehensive assessment
Local Linearization Formula
L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) \boxed{L(x) = f(a) + f'(a)(x - a)} L ( x ) = f ( a ) + f โฒ ( a ) ( x โ
Component Meaning a a a Base point (choose a "nice" value) f ( a ) f(a) f ( a ) Known function value at a a a f โฒ ( a ) f'(a) f
Worked Example
Approximate 4.1 \sqrt{4.1} 4.1 โ using linearization.
f ( x ) = x f(x) = \sqrt{x} f ( x ) = x โ , a = 4 a = 4 a = 4 (nearest perfect square).
f ( 4 ) = 2 f(4) = 2 f ( 4 ) = 2 , f โฒ ( x ) = 1 2 x f'(x) = \frac{1}{2\sqrt{x}} f โฒ ( x ) = 2 ,
L ( x ) = 2 + 1 4 ( x โ 4 ) L(x) = 2 + \frac{1}{4}(x-4) L ( x ) = 2 + 4 1 โ ( x โ 4 )
Actual: 4.1 โ 2.02485 \sqrt{4.1} \approx 2.02485 4.1 โ โ 2.02485 . Error โ 0.00015 \approx 0.00015 โ 0.00015 !
Key Fact: Choose a a a to be a nearby value where f ( a ) f(a) f ( a ) and f โฒ ( a ) f'(a) f โฒ ( a ) are easy to compute. The closer x x x is to , the better the approximation.
Practice โ Linearization ๐ฏ
Build the linearization. ๐
Key Takeaways โ Part 1
Linearization: L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) L(x) = f(a) + f'(a)(x-a) L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a )
Choose near where and are easy
Part 2: Differentials Linearization & Differentials
Part 2 of 7 โ Approximating Values
Common Linearizations at a = 0 a = 0 a = 0
Function Linear Approximation near 0 0 0 sin โก x \sin x sin x โ x \approx x โ
Part 3: Over/Underestimates Linearization & Differentials
Part 3 of 7 โ Differentials
Differentials vs. Derivatives
d y = f โฒ ( x ) โ d x \boxed{dy = f'(x)\,dx} d y = f โฒ ( x ) d x โ
Part 4: Percentage Error Linearization & Differentials
Part 4 of 7 โ Error Analysis
Error Terminology
Term Symbol Formula Approximate change d y dy d y f โฒ ( x ) โ d x f'(x)\,dx f โฒ ( x ) d x Exact change
Part 5: Linearization with Tables Linearization & Differentials
Part 5 of 7 โ Applications
Propagation of Error
If a measurement x x x has uncertainty d x dx d x , then the uncertainty in f ( x ) f(x) f ( x ) is approximately:
d f = โฃ f โฒ ( x ) โฃ โ d x \boxed{df = |f'(x)|\,dx}
Part 6: Problem-Solving Workshop Linearization & Differentials
Part 6 of 7 โ AP-Style Workshop
AP FRQ Patterns
Pattern What They Ask Table + tangent line "Use the tangent line at x = a x = a x = a to approximate f ( b ) f(b) f ( b ) " Over/under "Is your estimate an over- or underestimate? Justify." Differential "What is d y dy when and ?"
Part 7: Final Assessment Linearization & Differentials
Part 7 of 7 โ Comprehensive Assessment
Formula Reference
Formula Expression Linearization L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) L(x) = f(a) + f'(a)(x-a) L ( x ) = f ( a ) + f
a
)
โ
โฒ
(
a
)
Slope of tangent line at a a a
x โ a x - a x โ a Small displacement from a a a
1
โ
f โฒ ( 4 ) = 1 4 f'(4) = \frac{1}{4} f โฒ ( 4 ) = 4 1 โ L ( 4.1 ) = 2 + 1 4 ( 0.1 ) = 2.025 L(4.1) = 2 + \frac{1}{4}(0.1) = \boxed{2.025} L ( 4.1 ) = 2 + 4 1 โ ( 0.1 ) = 2.025 โ
a
a
The approximation improves as x โ a x \to a x โ a This is the tangent line at x = a x = a x = a used as an approximation
x
cos โก x \cos x cos x โ 1 \approx 1 โ 1
tan โก x \tan x tan x โ x \approx x โ x
e x e^x e x โ 1 + x \approx 1 + x โ 1 + x
ln โก ( 1 + x ) \ln(1+x) ln ( 1 + x ) โ x \approx x โ x
( 1 + x ) n (1+x)^n ( 1 + x ) n โ 1 + n x \approx 1 + nx โ 1 + n x
Key Fact: These linearizations appear frequently on the AP exam, especially e x โ 1 + x e^x \approx 1+x e x โ 1 + x and sin โก x โ x \sin x \approx x sin x โ x for small x x x .
Over/Under Estimates from Concavity Concaveย up โ Tangentย lineย underestimates \boxed{\text{Concave up} \Rightarrow \text{Tangent line underestimates}} Concaveย up โ Tangentย lineย underestimates โ
Concaveย down โ Tangentย lineย overestimates \boxed{\text{Concave down} \Rightarrow \text{Tangent line overestimates}} Concaveย down โ Tangentย lineย overestimates โ
Concavity at a a a Tangent line is a... f โฒ โฒ ( a ) > 0 f''(a) > 0 f โฒโฒ ( a ) > 0 (concave up)Underestimate f โฒ โฒ ( a ) < 0 f''(a) < 0 f โฒโฒ ( a ) < 0 (concave down)Overestimate
Worked Example
Approximate e 0.1 e^{0.1} e 0.1 using linearization. Is it an over- or underestimate?
f ( x ) = e x f(x) = e^x f ( x ) = e x , a = 0 a = 0 a = 0 . L ( x ) = 1 + x L(x) = 1 + x L ( x ) = 1 + x . L ( 0.1 ) = 1.1 L(0.1) = 1.1 L ( 0.1 ) = 1.1 .
f โฒ โฒ ( x ) = e x > 0 f''(x) = e^x > 0 f โฒโฒ ( x ) = e x > 0 everywhere โ concave up โ so tangent line is an underestimate .
Actual: e 0.1 โ 1.10517 e^{0.1} \approx 1.10517 e 0.1 โ 1.10517 . Indeed 1.1 < 1.10517 1.1 < 1.10517 1.1 < 1.10517 .
Practice โ Approximations ๐ฏ
Classify each approximation. ๐
Key Takeaways โ Part 2
Memorize common linearizations: sin โก x โ x \sin x \approx x sin x โ x , e x โ 1 + x e^x \approx 1+x e x โ 1 + x , ln โก ( 1 + x ) โ x \ln(1+x) \approx x ln ( 1 + x ) โ x
Concave up (f โฒ โฒ > 0 f'' > 0 f โฒโฒ > 0 ) โ \Rightarrow โ underestimate
Concave down (f โฒ โฒ < 0 f'' < 0 f โฒโฒ < 0 ) โ \Rightarrow โ overestimate
AP frequently asks "is this an over- or underestimate?"
Concept Notation Meaning Derivative d y d x = f โฒ ( x ) \frac{dy}{dx} = f'(x) d x d y โ = f โฒ ( x ) Instantaneous rate of change Differential of y y y d y = f โฒ ( x ) โ d x dy = f'(x)\,dx d y = f โฒ ( x ) d x Approximate change in y y y Actual change ฮ y = f ( x + ฮ x ) โ f ( x ) \Delta y = f(x+\Delta x) - f(x) ฮ y = f ( x + ฮ x ) โ f ( x ) Exact change in y y y
Relationship ฮ y โ d y whenย d x = ฮ x ย isย small \Delta y \approx dy \quad \text{when } dx = \Delta x \text{ is small} ฮ y โ d y whenย d x = ฮ x ย isย small
The differential d y dy d y is the change along the tangent line . The actual change ฮ y \Delta y ฮ y is the change along the curve .
Worked Example
y = x 3 y = x^3 y = x 3 . Find d y dy d y when x = 2 x = 2 x = 2 and d x = 0.01 dx = 0.01 d x = 0.01 .
d y = f โฒ ( x ) โ d x = 3 x 2 โ d x = 3 ( 4 ) ( 0.01 ) = 0.12 dy = f'(x)\,dx = 3x^2\,dx = 3(4)(0.01) = 0.12 d y = f โฒ ( x ) d x = 3 x 2 d x = 3 ( 4 ) ( 0.01 ) = 0.12
Actual change: ฮ y = ( 2.01 ) 3 โ 8 = 8.120601 โ 8 = 0.120601 \Delta y = (2.01)^3 - 8 = 8.120601 - 8 = 0.120601 ฮ y = ( 2.01 ) 3 โ 8 = 8.120601 โ 8 = 0.120601
d y = 0.12 dy = 0.12 d y = 0.12 vs ฮ y โ 0.1206 \Delta y \approx 0.1206 ฮ y โ 0.1206 โ very close!
Key Fact: d y dy d y is a linear approximation to ฮ y \Delta y ฮ y . The smaller d x dx d x , the better the approximation.
Practice โ Differentials ๐ฏ
Compare d y dy d y and ฮ y \Delta y ฮ y . ๐
Compute the differential. โ๏ธ
Key Takeaways โ Part 3
d y = f โฒ ( x ) โ d x dy = f'(x)\,dx d y = f โฒ ( x ) d x is the differential (change along tangent)
ฮ y = f ( x + d x ) โ f ( x ) \Delta y = f(x+dx) - f(x) ฮ y = f ( x + d x ) โ f ( x ) is the actual change
d y โ ฮ y dy \approx \Delta y d y โ ฮ y for small d x dx d x
The derivative d y / d x dy/dx d y / d x is the ratio of differentials
f ( x + ฮ x ) โ f ( x ) f(x+\Delta x) - f(x) f ( x + ฮ x ) โ f ( x )
Absolute error $ \Delta y - dy
Relative error $\frac{ \Delta y - dy
Percent error $\frac{ \Delta y - dy
Over/Under with Error Bounds Error โ 1 2 f โฒ โฒ ( a ) ( x โ a ) 2 \boxed{\text{Error} \approx \frac{1}{2}f''(a)(x-a)^2} Error โ 2 1 โ f โฒโฒ ( a ) ( x โ a ) 2 โ
This is the next term in the Taylor expansion. For small ( x โ a ) (x-a) ( x โ a ) , the error is approximately quadratic in the displacement.
Worked Example
Approximate 26 \sqrt{26} 26 โ using a = 25 a = 25 a = 25 . Estimate the error.
L ( 26 ) = 5 + 1 10 ( 1 ) = 5.1 L(26) = 5 + \frac{1}{10}(1) = 5.1 L ( 26 ) = 5 + 10 1 โ ( 1 ) = 5.1
Error โ 1 2 f โฒ โฒ ( 25 ) ( 1 ) 2 = 1 2 ( โ 1 4 โ
125 ) = โ 1 1000 = โ 0.001 \approx \frac{1}{2}f''(25)(1)^2 = \frac{1}{2}\left(-\frac{1}{4 \cdot 125}\right) = -\frac{1}{1000} = -0.001 โ 2 1 โ f โฒโฒ ( 25 ) ( 1 ) 2 = 2 1 โ ( โ 4 โ
125 1 โ ) = โ 1000 1 โ = โ 0.001
So 26 โ 5.1 โ 0.001 = 5.099 \sqrt{26} \approx 5.1 - 0.001 = 5.099 26 โ โ 5.1 โ 0.001 = 5.099 . Actual: 5.0990 โฆ 5.0990\ldots 5.0990 โฆ โ extremely close!
AP Tip: The sign of f โฒ โฒ ( a ) f''(a) f โฒโฒ ( a ) tells you whether L L L overestimates (f โฒ โฒ < 0 f'' < 0 f โฒโฒ < 0 ) or underestimates (f โฒ โฒ > 0 f'' > 0 f โฒโฒ > 0 ). The magnitude tells you how large the error is.
Practice โ Error Analysis ๐ฏ
Error classification. ๐
Calculate the error. โ๏ธ
Key Takeaways โ Part 4
Error in linearization โ ( x โ a ) 2 \propto (x-a)^2 โ ( x โ a ) 2 (quadratic)
Concave up โ \Rightarrow โ underestimate; concave down โ \Rightarrow โ overestimate
Absolute error = โฃ ฮ y โ d y โฃ = |\Delta y - dy| = โฃฮ y โ d y โฃ ; relative error = โฃ ฮ y โ d y โฃ / โฃ ฮ y โฃ = |\Delta y - dy|/|\Delta y| = โฃฮ y โ d y โฃ/โฃฮ y โฃ
Stay close to a a a for smaller errors
df = โฃ f โฒ ( x ) โฃ d x โ
Worked Example: Sphere Volume
A sphere has radius r = 5 r = 5 r = 5 cm with measurement error d r = 0.02 dr = 0.02 d r = 0.02 cm. Estimate the error in the volume.
V = 4 3 ฯ r 3 V = \frac{4}{3}\pi r^3 V = 3 4 โ ฯ r 3 , V โฒ ( r ) = 4 ฯ r 2 V'(r) = 4\pi r^2 V โฒ ( r ) = 4 ฯ r 2
d V = 4 ฯ ( 25 ) ( 0.02 ) = 2 ฯ โ 6.28 ย cm 3 dV = 4\pi(25)(0.02) = 2\pi \approx 6.28 \text{ cm}^3 d V = 4 ฯ ( 25 ) ( 0.02 ) = 2 ฯ โ 6.28 ย cm 3
Relative error: d V V = 4 ฯ r 2 โ d r 4 3 ฯ r 3 = 3 โ d r r = 3 ( 0.02 ) 5 = 0.012 = 1.2 % \frac{dV}{V} = \frac{4\pi r^2\,dr}{\frac{4}{3}\pi r^3} = \frac{3\,dr}{r} = \frac{3(0.02)}{5} = 0.012 = 1.2\% V d V โ = 3 4 โ ฯ r 3 4 ฯ r 2 d r โ = r 3 d r โ = 5 3 ( 0.02 ) โ = 0.012 = 1.2%
Key Fact: For V = 4 3 ฯ r 3 V = \frac{4}{3}\pi r^3 V = 3 4 โ ฯ r 3 , the relative error in volume is 3 times the relative error in radius: d V V = 3 d r r \frac{dV}{V} = 3\frac{dr}{r} V d V โ = 3 r d r โ .
Linearization from a Table AP problems often give a table of f f f and f โฒ f' f โฒ values and ask you to approximate f f f at a nearby point:
x x x f ( x ) f(x) f ( x ) f โฒ ( x ) f'(x) f โฒ ( x ) 2 2 2 5 5 5 3 3 3 4 4 4 11 11 11 โ 1 -1 โ 1
Approximate f ( 2.1 ) f(2.1) f ( 2.1 ) : L ( 2.1 ) = f ( 2 ) + f โฒ ( 2 ) ( 0.1 ) = 5 + 3 ( 0.1 ) = 5.3 L(2.1) = f(2) + f'(2)(0.1) = 5 + 3(0.1) = 5.3 L ( 2.1 ) = f ( 2 ) + f โฒ ( 2 ) ( 0.1 ) = 5 + 3 ( 0.1 ) = 5.3
Approximate f ( 3.9 ) f(3.9) f ( 3.9 ) : L ( 3.9 ) = f ( 4 ) + f โฒ ( 4 ) ( โ 0.1 ) = 11 + ( โ 1 ) ( โ 0.1 ) = 11.1 L(3.9) = f(4) + f'(4)(-0.1) = 11 + (-1)(-0.1) = 11.1 L ( 3.9 ) = f ( 4 ) + f โฒ ( 4 ) ( โ 0.1 ) = 11 + ( โ 1 ) ( โ 0.1 ) = 11.1
Practice โ Applications ๐ฏ
Propagation of error. โ๏ธ
Key Takeaways โ Part 5
Error propagation: d f โ โฃ f โฒ ( x ) โฃ โ d x df \approx |f'(x)|\,dx df โ โฃ f โฒ ( x ) โฃ d x
Relative error: d f / f df/f df / f . For V = c r n V = cr^n V = c r n : d V / V = n โ d r / r dV/V = n\,dr/r d V / V = n d r / r
Table-based linearization: L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) L(x) = f(a) + f'(a)(x-a) L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) using given values
AP loves "approximate f ( 2.1 ) f(2.1) f ( 2.1 ) given f ( 2 ) f(2) f ( 2 ) and f โฒ ( 2 ) f'(2) f โฒ ( 2 ) " questions
d
y
Setup "Write the linearization of f f f at x = a x = a x = a "
Full Worked AP Problem
The table below gives values of a twice-differentiable function f f f .
x x x 0 0 0 1 1 1 2 2 2 3 3 3 f ( x ) f(x) f ( x ) 4 4 4 7 7 7 3 3 3 1 1 1 f โฒ
(a) Write the linearization of f f f at x = 2 x = 2 x = 2 .
(b) Use your answer to approximate f ( 1.8 ) f(1.8) f ( 1.8 ) .
(c) Given f โฒ โฒ ( 2 ) = โ 1 f''(2) = -1 f โฒโฒ ( 2 ) = โ 1 , is your estimate an over- or underestimate? Justify.
Solution (a): L ( x ) = f ( 2 ) + f โฒ ( 2 ) ( x โ 2 ) = 3 + ( โ 3 ) ( x โ 2 ) = 3 โ 3 ( x โ 2 ) L(x) = f(2) + f'(2)(x-2) = 3 + (-3)(x-2) = 3 - 3(x-2) L ( x ) = f ( 2 ) + f โฒ ( 2 ) ( x โ 2 ) = 3 + ( โ 3 ) ( x โ 2 ) = 3 โ 3 ( x โ 2 )
Solution (b): L ( 1.8 ) = 3 โ 3 ( 1.8 โ 2 ) = 3 โ 3 ( โ 0.2 ) = 3 + 0.6 = 3.6 L(1.8) = 3 - 3(1.8-2) = 3 - 3(-0.2) = 3 + 0.6 = 3.6 L ( 1.8 ) = 3 โ 3 ( 1.8 โ 2 ) = 3 โ 3 ( โ 0.2 ) = 3 + 0.6 = 3.6
Solution (c): Since f โฒ โฒ ( 2 ) = โ 1 < 0 f''(2) = -1 < 0 f โฒโฒ ( 2 ) = โ 1 < 0 , f f f is concave down near x = 2 x = 2 x = 2 . The tangent line lies above the curve, so 3.6 3.6 3.6 is an overestimate .
AP Tip: For "justify," you must state the concavity (f โฒ โฒ < 0 f'' < 0 f โฒโฒ < 0 ), explain what it means (concave down), and conclude (tangent above curve = overestimate).
Justify your reasoning. ๐
Key Takeaways โ Part 6
AP FRQs often combine table data with linearization
Always justify over/underestimate with concavity
Three-part justification: sign of f โฒ โฒ f'' f โฒโฒ , concavity, conclusion
The tangent line approximation is exact when f f f is linear
โฒ
(
a
)
(
x
โ
a )
Differential d y = f โฒ ( x ) โ d x dy = f'(x)\,dx d y = f โฒ ( x ) d x
Error estimate โ 1 2 f โฒ โฒ ( a ) ( x โ a ) 2 \approx \frac{1}{2}f''(a)(x-a)^2 โ 2 1 โ f โฒโฒ ( a ) ( x โ a ) 2
Concave up (f โฒ โฒ > 0 f'' > 0 f โฒโฒ > 0 ) Tangent underestimates
Concave down (f โฒ โฒ < 0 f'' < 0 f โฒโฒ < 0 ) Tangent overestimates
Common Linearizations at a = 0 a = 0 a = 0 f ( x ) f(x) f ( x ) L ( x ) L(x) L ( x ) sin โก x \sin x sin x x x x cos โก x \cos x cos x 1 1 1 e x e^x e x 1 + x 1 + x 1 + x ln โก ( 1 + x ) \ln(1+x) ln ( 1 + x ) x x x ( 1 + x ) n (1+x)^n ( 1 + x ) n 1 + n x 1 + nx 1 + n x
Common AP Mistakes Mistake Correct Approach Choosing a a a far from x x x Pick a a a as close as possible Forgetting f โฒ ( a ) f'(a) f โฒ ( a ) in L ( x ) L(x) L ( x ) L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) L(x) = f(a) + f'(a)(x-a) L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) , not Incomplete justification State f โฒ โฒ f'' f โฒโฒ sign, concavity, AND over/under Confusing d y dy d y and ฮ y \Delta y ฮ y d y dy d y = tangent line change; ฮ y \Delta y ฮ y = actual change
Quiz Set 1 โ Core Skills ๐ฏ
Quiz Set 2 โ Applications ๐ฏ
๐ Topic Complete!
You've mastered Linearization & Differentials :
Part Topic Status 1 Tangent line approximation โ
2 Approximating values โ
3 Differentials โ
4 Error analysis โ
5 Applications โ
6 AP-style workshop โ
7 Comprehensive assessment โ
Key Fact: Linearization L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) L(x) = f(a) + f'(a)(x-a) L ( x ) = f ( a ) + f โฒ ( a ) ( x โ a ) is the foundation for approximation in calculus. On the AP exam, always pair your estimate with "over/underestimate" justified by concavity.
( x ) f'(x) f โฒ ( x )
f ( a ) + ( x โ a ) f(a) + (x-a) f ( a ) + ( x โ a )