Calculus Solutions EX # 3.5

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Summary: Local linear appronimation of f at x_{0} :
Local linear appronimation of f at x_{0} :

    \[f(x) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-----(1)\]

Let \Delta x=x-x_{0} then

    \[f\left(x_{0}+\Delta x\right) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right) \Delta x------(2)\]

Differentials:

    \[\begin{aligned} & d y=f^{\prime}(x) d x ----------(5)\\ & f^{\prime}(x)=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \\ \\ & f^{\prime}(x) \approx \frac{\Delta y}{\Delta x} \\ \\ & \Delta y \approx f^{\prime}(x) \Delta x \\ \\ & \text { Let } \Delta x=d x \text { then } \\ \\ & \Delta y \approx f^{\prime}(x) d x = d y ---------(7) \\ \\ & \text {at}\quad x=x_{0} \\ \\ & \Delta y \approx dy=f^{\prime}(x_{0}) d x = d y ---------(8) \\ \\ & \Delta y \approx dy=f^{\prime}(x) d x = d y ---------(9) \end{aligned}\]

Measurement errer of x :

    \[d x(=\Delta x)=x-x_{0}\]

Propagated error of y :

    \[\Delta y=f(x)-f\left(x_{0}\right)\]

Q.1 (a) use formula (1) to obtain the local linear approximation of x^3 at x_{0}=1. (b) Use formula (2) to rewrite the approximation obtained in part (a) interms of \Delta x
Solution:
(a)

    \[\begin{aligned} & f(x)=x^{3} ; \\ \\ & f^{\prime}(x)=3 x^{2} \\ \\ & f^{\prime}(1)=3(1)^{2}=3 \end{aligned}\]

    \[\begin{aligned} & f(x)=f(1)+f^{\prime}(1)(x-1) \\ \\ & f(x) \approx 1+3(x-1) \end{aligned}\]

(b)

    \[\begin{aligned} & f\left(x_{0}+\Delta x\right)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right) \Delta x ----(1)\\ \\ & x_{0}=1, \quad f(x)=x^{3}, \quad f(1)=1 \\ \\& f^{\prime}(x)=3 x^{2} \Rightarrow f^{\prime}(1)=3(1)^{2}=3 \\ \\ & \Delta x=x-x_{0}=x-1 \\ \\ & \therefore (1) \Rightarrow f(1+\Delta x)=1+3 \Delta x \end{aligned}\]

(C)

    \[(1.02)^{3}\]

    \[\begin{aligned} & \text {from part (a)} \quad f(x) \approx 1+3(x-1) \\ \\ & \text { Put } x=1.02 \\ \\ & (1.02)^{3} \approx (1+3(1.02-1)=1+3(0.02)=1+0.06 =1.06\\ \\ & \text { from part (b) } \\ \\ & f(1+\Delta x)=1+3 \Delta x \\ \\ & (1.02)^{3}=(1+0.02)^{3} \\ \\ & \Delta x=0.02 \\ \\ & \therefore \quad(1.02)^{3} \approx 1+3(.02)=1+.06=1.06 \end{aligned}\]

Q.2 (a) use formula (1) to obtain the local linear approximation of \frac{1}{x} at x_{0}=2 . (b) Use formula (2) to rewrite the approximation obtained in part (a) interms of \Delta x
Solution:
(a)

    \[\begin{aligned} & f(x)=\frac{1}{x} ; \quad x_{0}=2, \quad f(2)=\frac{1}{2} \\ \\ & f^{\prime}(x)=-\frac{1}{x^{2}}, \quad f^{\prime}(2)=-\frac{1}{2^{2}}=-\frac{1}{4} \\ \\ & f(x) \approx f(2)+f^{\prime}(2)(x-2) \\ \\ & f(x)=\frac{1}{2}-\frac{1}{4}(x-2) \end{aligned}\]

(b)

    \[\begin{aligned} & f(x)=\frac{1}{x}, \quad x_{0}=2, \quad f(2)=1 / 2 \\ \\ & f^{\prime}(x)=-\frac{1}{x^{2}}, \quad f^{\prime}(2)=- \frac{1}{4}\\ \\ & f\left(x_{0}+\Delta x\right) \approx f\left(x_{0}\right)+f^{\prime}(x_{0}) \Delta x \\ \\ & f\left(x_{0}+\Delta x\right) \approx \frac{1}{2}-\frac{1}{4}\Delta x \end{aligned}\]

(c)

    \[\frac{1}{ 2.05} \Rightarrow x=2.05, \quad x_{0}=2, \quad \Delta x=0.05\]

From part(a)

    \[\begin{aligned} \frac{1}{2.05} \approx \frac{1}{2}-\frac{1}{4}(2.05-2) & =0.5-0.25(.05) \\ \\ & =0.5-0.0125=0.4875 \end{aligned}\]

From part (b)

    \[\begin{aligned} f(2+\Delta x) & \approx \frac{1}{2}-\frac{1}{4} \Delta x \\ \\ \frac{1}{2.05} & \approx 0.5-0.25(0.05)=0.4875 \end{aligned}\]

Q.5-6 confirm that the stated formula is the local linear approximation at x_{0}=0
Solution:

    \[\begin{aligned} & (1+x)^{15} \approx 1+15 x \\ \\ & f(x)=(1+x)^{15}, \quad x_{0}=1 . \quad f(0)=(1+0)^{15}=1 \\ \\ & f^{\prime}(x)=15(1+x)^{14} ; \quad f^{\prime}(0)=15(1+0)^{14}=15 \\ \\ & f(x) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) \\ \\ & f(x) \approx f(0)+f^{\prime}(0)(x-0) \\ \\ & f(x) \approx 1+15 x \end{aligned}\]

6.

    \[\begin{aligned} & \frac{1}{\sqrt{1-x}} \approx 1+\frac{1}{2} x \\ \\ & f(x)=\frac{1}{\sqrt{1-x}}=(1-x)^{-\frac{1}{2}} ; \quad x_{0}=0, \\ \\ & f(0)=\frac{1}{\sqrt{1-0}}=1\right. \\ \\ & f^{\prime}(x)=-\frac{1}{2} ;(1-x)^{\frac{1}{2}-1}(-1) \\ \\ & f^{\prime}(x)=\frac{1}{2}(1-x)^{-3 / 2} ; \quad f^{\prime}(0)=\frac{1}{2}(1-0)^{-3 / 2}=\frac{1}{2} \\ \\ & f(x) \approx f\left(x_{0}\right)+f^{\prime}(x)\left(x-x_{0}\right) \\ \\ & \frac{1}{\sqrt{1-x}} \approx f(0)+f^{\prime}(0)(x-0) \\ & \frac{1}{\sqrt{1-x}} \approx 1+\frac{1}{2} x \end{aligned}\]