Answer to Question #144709 in Calculus for Besmallah Yousefi

Question

1. What is the maximum error possible in using the Taylor polynomial approximation

sin(x)≈x-((x^3)/3!)+((x^5)/5!) when -0.3≤x≤0.3?. Use this approximation to find sin(12^0) correct to six decimal places. (Note that 12^0=12π/180).

1. Integrate the function f(x,y)=6x+(2y^2) over the region R bounded by the parabola x=y^2 and the straight line x+y=2.

Answer 1

1.The remainder for a Taylor series is calculated as:

$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}$ , where c is from the interval of approximation.

So the maximum error is calculated as:

$|R_n(x)|\leq\frac{M}{(n+1)!}|x-a|^{(n+1)}$ , where M is a maximum value of $|f^{(n+1)}(x)|$ on the interval.

In our case $\frac{x^5}{5}$ is a fifth, so n=5. a=0. f(x)=sin(x), f'(x)=cos(x), f''(x)=-sin(x), f'''(x)=-cos(x), f''''(x)=sin(x).

$f^{(5)}(x)=cos(x), f^{(6)}(x)=-sin(x)$ . sin is rising from -0.3 to 0.3 and sin(0.3)=-sin(-0.3), so M=sin(0.3) and we get the maximum error (where -0.3<x<0.3):

$\frac{M}{(n+1)!}|x-a|^{(n+1)}=\frac{sin(0.3)}{6!}|x|^{(6)}\approx0.00041|x|^6$ .

Let's calculate using this approximation:

$sin(\frac{12\pi}{180})=\frac{12\pi}{180}-\frac{12^3\pi^3}{180^3\cdot3!}+\frac{12^5\pi^5}{180^5\cdot5!}=0.207912$ .

2.Let's find out where do $x=y^2$ and $x=2-y$ intercept:

$y^2=2-y$

$y^2+y-2=0$

$y_1=-2, y_2=1$

For every y from -2 to 1 we have $y^2\leq2-y$ (we just need to check in one point, let's say y=0), so the area is bounded by the line x=2-y in the top and the line $x=y^2$ in the bottom and y varies from -2 to 1.

$\int_{-2}^{1}dy\int_{y^2}^{2-y}(6x+2y^2) dx=\int_{-2}^{1}(3x^2+2y^2x)|_{y^2}^{2-y}dy=$

$=\int_{-2}^1(3\cdot(2-y)^2+2y^2(2-y)-3y^4-2y^4)dy=$

$=\int_{-2}^1(12-12y+3y^2+4y^2-2y^3-5y^4)dy=$

$=\int_{-2}^1(-5y^4-2y^3+7y^2-12y+12)dy=$

$=(-y^5-\frac{y^4}{2}+\frac{7y^3}{3}-6y^2+12y)|_{-2}^1=$

$=-1-\frac{1}{2}+\frac{7}{3}-6+12-32+8+\frac{7\cdot8}{3}+24+24=\frac{103}{2}=51.5$

Answer is 51.5