Linear Approximations and Differentials

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find the linear approximation of \(f(x)=\sqrt{x}\) at \(x=9\) and use the approximation to estimate \(\sqrt{9.1}.\)

   i

   Since we are looking for the linear approximation at \(x=9,\) using we know the linear approximation is given by

   \[L(x)=f(9)+f'(9)(x-9).\]

   We need to find \(f(9)\) and \(f'(9).\)

   \[\begin{array}{lll}f(x)=\sqrt{x} & ⇒ & f(9)=\sqrt{9}=3 \\ f'(x)=\frac{1}{2\sqrt{x}} & ⇒ & f'(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\end{array}\]

   Therefore, the linear approximation is given by .

   \[L(x)=3+\frac{1}{6}(x-9)\]

   Using the linear approximation, we can estimate \(\sqrt{9.1}\) by writing

   \[\sqrt{9.1}=f(9.1)\approx L(9.1)=3+\frac{1}{6}(9.1-9)\approx 3.0167.\]

2. Find the local linear approximation to \(f(x)=\sqrt[3]{x}\) at \(x=8.\) Use it to approximate \(\sqrt[3]{8.1}\) to five decimal places.

   i

   \(L(x)=2+\frac{1}{12}(x-8);\) 2.00833

3. Find the linear approximation of \(f(x)=\text{sin}\ x\) at \(x=\frac{\pi }{3}\) and use it to approximate \(\text{sin}(62\text{^{\circ}}).\)

   i

   First we note that since \(\frac{\pi }{3}\) rad is equivalent to \(60\text{^{\circ}},\) using the linear approximation at \(x=\pi \text{/}3\) seems reasonable. The linear approximation is given by

   \[L(x)=f(\frac{\pi }{3})+f'(\frac{\pi }{3})(x-\frac{\pi }{3}).\]

   We see that

   \[\begin{array}{lll}f(x)=\text{sin}\ x & ⇒ & f(\frac{\pi }{3})=\text{sin}(\frac{\pi }{3})=\frac{\sqrt{3}}{2} \\ f'(x)=\text{cos}\ x & ⇒ & f'(\frac{\pi }{3})=\text{cos}(\frac{\pi }{3})=\frac{1}{2}\end{array}\]

   Therefore, the linear approximation of \(f\) at \(x=\pi \text{/}3\) is given by .

   \[L(x)=\frac{\sqrt{3}}{2}+\frac{1}{2}(x-\frac{\pi }{3})\]

   To estimate \(\text{sin}(62\text{^{\circ}})\) using \(L,\) we must first convert \(62\text{^{\circ}}\) to radians. We have \(62\text{^{\circ}}=\frac{62\pi }{180}\) radians, so the estimate for \(\text{sin}(62\text{^{\circ}})\) is given by

   \[\text{sin}(62\text{^{\circ}})=f(\frac{62\pi }{180})\approx L(\frac{62\pi }{180})=\frac{\sqrt{3}}{2}+\frac{1}{2}(\frac{62\pi }{180}-\frac{\pi }{3})=\frac{\sqrt{3}}{2}+\frac{1}{2}(\frac{2\pi }{180})=\frac{\sqrt{3}}{2}+\frac{\pi }{180}\approx 0.88348.\]

4. Find the linear approximation for \(f(x)=\text{cos}\ x\) at \(x=\frac{\pi }{2}.\)

   i

   \(L(x)=\text{-}x+\frac{\pi }{2}\)

5. Find the linear approximation of \(f(x)={(1+x)}^{n}\) at \(x=0.\) Use this approximation to estimate \({(1.01)}^{3}.\)

   i

   The linear approximation at \(x=0\) is given by

   \[L(x)=f(0)+f'(0)(x-0).\]

   Because

   \[\begin{array}{lll}f(x)={(1+x)}^{n} & ⇒ & f(0)=1 \\ f'(x)=n{(1+x)}^{n-1} & ⇒ & f'(0)=n,\end{array}\]

   the linear approximation is given by (a).

   \[L(x)=1+n(x-0)=1+nx\]

   We can approximate \({(1.01)}^{3}\) by evaluating \(L(0.01)\) when \(n=3.\) We conclude that

   \[{(1.01)}^{3}=f(1.01)\approx L(0.01)=1+3(0.01)=1.03.\]

6. Find the linear approximation of \(f(x)={(1+x)}^{4}\) at \(x=0\) without using the result from the preceding example.

   i

   \(L(x)=1+4x\)

7. For each of the following functions, find dy and evaluate when \(x=3\) and \(dx=0.1.\)

   1. \(y={x}^{2}+2x\)
   2. \(y=\text{cos}\ x\)
   i

   The key step is calculating the derivative. When we have that, we can obtain dy directly.

   1. Since \(f(x)={x}^{2}+2x,\) we know \(f'(x)=2x+2,\) and therefore
      \[dy=(2x+2)dx.\]
      When \(x=3\) and \(dx=0.1,\)
      \[dy=(2\cdot 3+2)(0.1)=0.8.\]
   2. Since \(f(x)=\text{cos}\ x,\) \(f'(x)=\text{-}\text{sin}(x).\) This gives us
      \[dy=\text{-}\text{sin}\ x\ dx.\]
      When \(x=3\) and \(dx=0.1,\)
      \[dy=\text{-}\text{sin}(3)(0.1)=-0.1\ \text{sin}(3).\]

8. For \(y={e}^{{x}^{2}},\) find \(dy.\)

   i

   \(dy=2x{e}^{{x}^{2}}dx\)

9. Let \(y={x}^{2}+2x.\) Compute \(\text{\Delta }y\) and dy at \(x=3\) if \(dx=0.1.\)

   i

   The actual change in \(y\) if \(x\) changes from \(x=3\) to \(x=3.1\) is given by

   \[\text{\Delta }y=f(3.1)-f(3)=[{(3.1)}^{2}+2(3.1)]-[{3}^{2}+2(3)]=0.81.\]

   The approximate change in \(y\) is given by \(dy=f'(3)dx.\) Since \(f'(x)=2x+2,\) we have

   \[dy=f'(3)dx=(2(3)+2)(0.1)=0.8.\]

10. For \(y={x}^{2}+2x,\) find \(\text{\Delta }y\) and \(dy\) at \(x=3\) if \(dx=0.2.\)

    i

    \(dy=1.6,\) \(\text{\Delta }y=1.64\)

11. Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.

    1. Use differentials to estimate the error in the computed volume of the cube.
    2. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.
    i

    1. The measurement of the side length is accurate to within \(\text{\pm }0.1\) cm. Therefore,
       \[-0.1\le dx\le 0.1.\]
       The volume of a cube is given by \(V={x}^{3},\) which leads to
       \[dV=3{x}^{2}dx.\]
       Using the measured side length of 5 cm, we can estimate that
       \[-3{(5)}^{2}(0.1)\le dV\le 3{(5)}^{2}(0.1).\]
       Therefore,
       \[-7.5\le dV\le 7.5.\]
    2. If the side length is actually 4.9 cm, then the volume of the cube is
       \[V(4.9)={(4.9)}^{3}=117.649{\ \text{cm}}^{3}.\]
       If the side length is actually 5.1 cm, then the volume of the cube is
       \[V(5.1)={(5.1)}^{3}=132.651{\ \text{cm}}^{3}.\]
       Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is \(V(5)={5}^{3}=125.\) Therefore, the error in the computed volume is
       \[117.649-125\le \text{\Delta }V\le 132.651-125.\]
       That is,
       \[-7.351\le \text{\Delta }V\le 7.651.\]
       We see the estimated error \(dV\) is relatively close to the actual potential error in the computed volume.

12. Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.

    i

    The volume measurement is accurate to within \(21.6{\ \text{cm}}^{3}.\)

13. An astronaut using a camera measures the radius of Earth as 4000 mi with an error of \(\text{\pm }80\) mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.

    i

    If the measurement of the radius is accurate to within \(\text{\pm }80,\) we have

    \[-80\le dr\le 80.\]

    Since the volume of a sphere is given by \(V=(\frac{4}{3})\pi {r}^{3},\) we have

    \[dV=4\pi {r}^{2}dr.\]

    Using the measured radius of 4000 mi, we can estimate

    \[-4\pi {(4000)}^{2}(80)\le dV\le 4\pi {(4000)}^{2}(80).\]

    To estimate the relative error, consider \(\frac{dV}{V}.\) Since we do not know the exact value of the volume \(V,\) use the measured radius \(r=4000\ \text{mi}\) to estimate \(V.\) We obtain \(V\approx (\frac{4}{3})\pi {(4000)}^{3}.\) Therefore the relative error satisfies

    \[\frac{-4\pi {(4000)}^{2}(80)}{4\pi {(4000)}^{3}\text{/}3}\le \frac{dV}{V}\le \frac{4\pi {(4000)}^{2}(80)}{4\pi {(4000)}^{3}\text{/}3},\]

    which simplifies to

    \[-0.06\le \frac{dV}{V}\le 0.06.\]

    The relative error is 0.06 and the percentage error is \(6\text{\%}.\)

14. Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of \(\text{\pm }100\) mi.

    i

    7.6%

15. What is the linear approximation for any generic linear function \(y=mx+b?\)

16. Determine the necessary conditions such that the linear approximation function is constant. Use a graph to prove your result.

    i

    \({f}^{'}(a)=0\)

17. Explain why the linear approximation becomes less accurate as you increase the distance between \(x\) and \(a.\) Use a graph to prove your argument.

18. When is the linear approximation exact?

    i

    The linear approximation exact when \(y=f(x)\) is linear or constant.

19. \(f(x)=x+{x}^{4},a=0\)

20. \(f(x)=\frac{1}{x},a=2\)

    i

    \(L(x)=\frac{1}{2}-\frac{1}{4}(x-2)\)

21. \(f(x)=\text{tan}\ x,a=\frac{\pi }{4}\)

22. \(f(x)=\text{sin}\ x,a=\frac{\pi }{2}\)

    i

    \(L(x)=1\)

23. \(f(x)=x\ \text{sin}\ x,a=2\pi\)

24. \(f(x)={\text{sin}}^{2}x,a=0\)

    i

    \(L(x)=0\)

25. [T] \({(2.001)}^{6}\)

26. [T] \(\text{sin}(0.02)\)

    i

    0.02

27. [T] \(\text{cos}(0.03)\)

28. [T] \({(15.99)}^{1\text{/}4}\)

    i

    \(1.9996875\)

29. [T] \(\frac{1}{0.98}\)

30. [T] \(\text{sin}(3.14)\)

    i

    \(0.001593\)

31. [T] \({(1.01)}^{3}\)

32. [T] \(\text{cos}(0.01)\)

    i

    \(1;\) error, \(\sim 0.00005\)

33. [T] \({(\text{sin}(0.01))}^{2}\)

34. [T] \({(1.01)}^{-3}\)

    i

    \(0.97;\) error, \(\sim 0.0006\)

35. [T] \({(1+\frac{1}{10})}^{10}\)

36. [T] \(\sqrt{8.99}\)

    i

    \(3-\frac{1}{600};\) error, \(\sim 4.632\ \times \ {10}^{-7}\)

37. \(y=3{x}^{4}+{x}^{2}-2x+1\)

38. \(y=x\ \text{cos}\ x\)

    i

    \(dy=(\text{cos}\ x-x\ \text{sin}\ x)dx\)

39. \(y=\sqrt{1+x}\)

40. \(y=\frac{{x}^{2}+2}{x-1}\)

    i

    \(dy=(\frac{{x}^{2}-2x-2}{{(x-1)}^{2}})dx\)

Kuri Gukoresha

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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