AP Calculus BC: 10.11 Finding Taylor Polynomial Approximations of Functions – Exam Style questions with Answer- FRQ

Question

Let y = f(x) be the particular solution to the differential equation \(\frac{dy}{dx}=y\cdot (x In x)\) with initial condition f(1) = 4 . It can be shown that f”(1) = 4.
(a) Write the second-degree Taylor polynomial for f about x = 1. Use the Taylor polynomial to approximate f(2) .
(b) Use Euler’s method, starting at x = 1 with two steps of equal size, to approximate f(2) . Show the work that leads to your answer.
(c) Find the particular solution y = f(x) to the differential equation \(\frac{dy}{dx}=y\cdot (x In x)\) with initial condition f(1) = 4 . 

Answer/Explanation

Ans:

(a)

\(\frac{dy}{dx}= 4.1 In 1\)

\(P_{2}(x)=4+In 1 (x-1)+\frac{4(x-1)^{2}}{2!}\)                                \(P_{2}(x)=4+\frac{4(x-1)^{2}}{2!}\)

\(P_{2}(2)=4+\frac{4(2-1)^{2}}{2!}= 4+2 = 6\)

(b)

\(\frac{dy}{dx}= 4\cdot 1.5 In 1.5\)

6 In 1.5

f(2) = 4 + 3 In 1.5

(c)

\(\frac{dy}{dx}= y\cdot sin x\)

\(\frac{1}{y}dy= x In x dx\)                                                           u = In x                                               dv = x dx

\(\int \frac{1}{y}dy= \int x In x dx\)                                            \(du = \frac{1}{x}dx\)                  \(v = \frac{1}{x}x^{2}\)

\(In y = \frac{1}{2}x^{2}In x -\frac{1}{4}x^{2}+ c\)

\(In 4 = \frac{1}{2}In 1 -\frac{1}{4}+ c\)

\(In 4 + \frac{1}{4}= c\)

\(In y =\frac{1}{2}x^{2}In x- \frac{1}{4}x^{2}+In 4 + \frac{1}{4}\)
\(y =4e^{\frac{1}{2}x^{2}In x}- \frac{1}{4}x^{2} + \frac{1}{4}\)

Question

A function f has derivatives of all orders for all real numbers x. A portion of the graph of f is shown above, along with the line tangent to the graph of f at x = 0. Selected derivatives of f at x = 0 are given in the table above.
(a) Write the third-degree Taylor polynomial for f about x = 0.
(b) Write the first three nonzero terms of the Maclaurin series for  ex . Write the second-degree Taylor polynomial for ex f(x) about x = 0.
(c) Let h be the function defined b \(h(x)=\int_{0}^{x}f(t)dt.\) Use the Taylor polynomial found in part (a) to find an approximation for h(1).
(d) It is known that the Maclaurin series for h converges to h (x) for all real numbers x. It is also known that the individual terms of the series for h(1 ) alternate in sign and decrease in absolute value to 0. Use the alternating series error bound to show that the approximation found in part (c) differs from h(1 ) by at most 0.45. 

Answer/Explanation

Ans:

(a) f (0) = 3 and f ‘(0) = -2
The third-degree Taylor polynomial for f about x = 0 is 

\(3-2x+\frac{3}{2!}x^{2}+\frac{-\frac{23}{2}}{3!}x^{3}=3-2x+\frac{3}{2}x^{2}-\frac{23}{12}x^{3}.\)

(b) The first three nonzero terms of the Maclaurin series for  ex are \(1+x+\frac{1}{2!}x^{2}.\)

The second-degree Taylor polynomial for ex f(x) about x = 0 is \(3\left ( 1+x+\frac{1}{2!}x^{2} \right )-2x(1+x)+\frac{3}{2!}x^{2}(1)\)

\(=3+(3-2)x+\left ( \frac{3}{2}-2+\frac{3}{2} \right )x^{2}\)

= 3 + x + x2.

(c) \(h(1)=\int_{0}^{1}f(t)dt\)

\(\approx \int_{0}^{1}\left ( 3-2t+\frac{3}{2}t^{2}-\frac{23}{12}t^{3} \right )dt\)

\(= \left [ 3t-t^{2}+\frac{1}{2}t^{3}-\frac{23}{48}t^{4} \right ]_{t=0}^{t=1}\)

\(=3-1+\frac{1}{2}-\frac{23}{48}=\frac{97}{48}\)

(d) The alternating series error bound is the absolute value of the first omitted term of the series for h(1).

\(\int_{0}^{1}\left ( \frac{54}{4!}t^{4} \right )dt=\left [ \frac{9}{20}t^{5} \right ]_{t=0}^{t=1}=\frac{9}{20}\)

\(Error \leq \left | \frac{9}{20} \right |=0.45\)

Question

A function f has derivatives of all orders at x = 0. Let Pn (x) denote the nth-degree Taylor polynomial for f about x = 0.
(a) It is known that f(0) = -4 and that \(p_{1}\left ( \frac{1}{2} \right )\) Show that f'(0) = 2.
(b) It is known that \(f”(0)=-\frac{2}{3}\) and \(f”‘(0)=\frac{1}{3}.\) Find p3(x).
(c) The function h has first derivative given by h'(x) = f(2x). It is known that h(0) = 7. Find the third-degree Taylor polynomial for h about x = 0.

Answer/Explanation

Ans:

(a) p1(x) = f(0) + f'(0)x = -4+f'(0)x

\(p_{1}\left ( \frac{1}{2} \right )=-4+f'(0)\cdot \frac{1}{2}=-3\)

\(f'(0)\cdot \frac{1}{2}=1\)

f'(0) = 2

(b) \(p_{3}(x)=-4+2x+\left ( -\frac{2}{3} \right )\cdot \frac{x^{2}}{2!}+\frac{1}{3}\cdot \frac{x^{3}}{3!}\)

\(=-4 + 2x -\frac{1}{3}x^{2}+\frac{1}{18}x^{3}\)

(c) Let Qn(x) denote the Taylor polynomial of degree n for h about x = 0.

\(h'(x)=f(2x)\Rightarrow Q’_{3}(x)=-4 + 2(2x) -\frac{1}{3}(2x)^{2}\)

\( Q_{3}(x)=-4x + 4\cdot \frac{x^{2}}{2}-\frac{4}{3}\cdot \frac{x^{3}}{3}+C; C=Q_{0}=h(0)=7\)

\( Q_{3}(x)=7-4x + 2x^{2}-\frac{4}{9}x^{3}\)

OR

h'(x) = f(2x), h”(x) = 2f'(2x), h”‘(x) = 4f”(2x)

\(h'(0) = f(0)=-4 h”(0) = 2f'(0)=4, h”‘(0) = 4f”(0)=-\frac{8}{3}\)

\(Q_{3}(x) = 7-4x+4\cdot \frac{x^{3}}{2!}-\frac{8}{3}\cdot \frac{x^{3}}{3!}=7-4x+2x^{2}-\frac{4}{9}x^{3}\)

Question

 (A) Approximate \(f(x)=\frac{1}{x^{2}}\) by \(T_{3}(x)\), the Taylor polynomial with degree 3 centered at 1.
(B) Use Taylor’s inequality to estimate the accuracy of the approximation \(f (x) ≈ T_{3}(x)\) when x lies in 0. 8≤ x ≤ 1 .2.

Answer/Explanation

Question

x11.11.21.31.4
f'(x)810121314.5

The function f is twice differentiable for x > 0 with f(1) = 15 and f”(1) = 20. Values of f’, the derivative of f, are given for selected values of x in the table above.
(a) Write an equation for the line tangent to the graph of f at x =1. Use this line to approximate f(1.4).
(b) Use a midpoint Riemann sum with two subintervals of equal length and values from the table to approximate \(\) Use the approximation for \(\) to estimate the value of f(1.4). Show
the computations that lead to your answer.
(c) Use Euler’s method, starting at x = 1 with two steps of equal size, to approximate f(1.4). Show the computations that lead to your answer.
(d) Write the second-degree Taylor polynomial for f about x = 1. Use the Taylor polynomial to approximate f(1.4).

Answer/Explanation

Ans:

(a) f(1) = 15, f'(1) = 8

An equation for the tangent line is
y  = 15 + 8 (x – 1) .

f(1.4) ≈ 15 + 8 (1.4 – 1) = 18.2

(b) \(\int_{1}^{1.4}f'(x)dx\approx (0.2)(10)+(0.2)(13)=4.6\)

\(f(1.4)=f(1)+\int_{1}^{1.4}f'(x)dx\)

f(1.4) ≈ 15 + 4.6 = 19.6

(c) f(1.2) ≈ f(1) + (0.2) (8) = 16.6

f(1.4) ≈ 16.6 + (0.2) (12) = 19.0

(d) \(T_{2}(x)=15+8(x-1)+\frac{20}{2!}(x-1)^{2}\)

= 15 + 8 (x-1) + 10(x – 1)2

f(1.4) ≈ 15 + 8(1.4-1) + 10(1.4 – 1)2 = 19.8

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