Explanation Given: ( x 2 + y 2 ) 2 = 25 3 ( x 2 − y 2 ) at ( 2 , 1 ) Calculation: Slope of the tangent is the derivative of the curve at the given point. So differentiating given curve with respect to x, we get ( x 2 + y 2 ) 2 = 25 3 ( x 2 − y 2 ) ⇒ d ( x 2 + y 2 ) 2 d x = 25 3 ⋅ d ( x 2 − y 2 ) d x ⇒ 2 ( x 2 + y 2 ) ⋅ d ( x 2 + y 2 ) d x = 25 3 [ d ( x 2 − y 2 ) d x ] ⇒ 2 ( x 2 + y 2 ) ⋅ ( 2 x + 2 y ⋅ d y d x ) = 25 3 [ 2 x − 2 y ⋅ d y d x ] [ ∵ d d x ( x n ) = n ⋅ x n − 1 ] ⇒ 2 ( x 2 + y 2 ) ⋅ ( x + y ⋅ d y d x ) = 25 3 [ x − y ⋅ d y d x ] ⇒ 2 x 3 + 2

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.6, Problem 32PS

To determine

The slope of the tangent to the given curve

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Chapter 3.6, Problem 31PS

Chapter 3.6, Problem 33PS

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

Could you explain this using the formula I attached and polar coordinates

Chapter 3 Solutions

Calculus

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Chapter 3 Solutions