Using linear approximation and tangent

Question

1.Let y=4x^2+2x+3. If Δx=0.3 at x=4, use linear approximation to estimate Δy

Δy≈

2.Let y=sin(3x). If Δx=0.3 at x=0, use linear approximation to estimate Δy

Δy≈ =   

Find the percentage error

error =  % 

3.Given the function below

f(x)=3√-18x^3+45

(3 goes in the top small part of the sqrt and -18x^3+45 all goes under/in the bigger part)

Find the equation of the tangent line to the graph of the function at x=1. Answer in mx+b form.

L(x)=  

Use the tangent line to approximate ƒ(1.1).

L(1.1)=

Compute the actual value of ƒ(1.1). What is the error between the function value and the linear approximation?

Answer as a positive value only.

|error|≈ (Approximate to at least 5 decimal places.)

4.Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.04 cm thick to a hemispherical dome with a diameter of 40 meters.

  ? cubic centimeters

Answer 1

At x=xo

f(x)=f(xo) + f'(xo)×(x - xo)

x - xo=delta x

Delta y=f'(xo)×(x - xo)

1) y=4x² + 2x + 3

f'(x)=8x + 2, delta x=0.3

f'(4)=8×4 +2 or 34

Delta y=34×0.3 or 10.2

2)

y=sin(3x)

f'(x)=3cos(3x), delta x=0.3

At x=0

f'(0)=3cos(0) or 3

Delta y=3×0.3 or 0.9

Y approximate of at x=0.3=f(0) + 0.9 or 0 + 0.9

Y approximate=0.9

Real f(0.3)=sin(3×0.3) or sin(0.9)

Real f(0.3)=0.783

Error=(0.9-0.783)/0.783 ×100%

Error=14.9%

3.

f(x)=(-18x³ + 45)^1/3

f'(x)=(1/3)*-54x²(-18x³ + 45)^-2/3

At x=1, f(1)=3, f'(1)=-2

L - 3=-2(x - 1)

L(x)=-2x + 5

L(1.1)=-2×1.1 + 5 or 2.9

Real f(1.1)=(-18(1.1)³ + 45)^1/3 or 2.76

error=(2.9-2.76)/2.76 ×100% or 5.04%

4.

Surface area=2*pi*r²

S'=4*pi*r, r=20 m

Delta r=0.0004 m

Delta S=4*pi×20 m*0.0004 or 0.1005 m²