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John Abbott · Engineering Technologies · 201-225-AB

201-225-AB Mathematical Models 2 tutoring

Work on derivatives, integration and differential equations for John Abbott Engineering Technologies. Start with the Newton’s-method example below, or bring a question to a one-to-one lesson, even if you have not started the calculation.

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Calculus for Engineering Technologies

Mathematical Models 2 continues Mathematical Models 1. John Abbott’s course description includes derivative applications, integration techniques, Fourier series and differential equations. The example below focuses on Newton’s method: using a tangent line to improve an approximate solution to an equation.

Topics to work on

  • Slopes and derivatives, including logarithmic and trigonometric functions
  • Newton’s method and minimum-and-maximum problems
  • Antiderivatives, definite integrals and area
  • Algebraic and trigonometric substitutions; integration by parts
  • Fourier series
  • Differential equations and separation of variables

Before trying the example

Review evaluating a polynomial, differentiating powers and working with signed fractions. You will use a function value and a slope at the same input. Keep the full intermediate value when calculating; round only the displayed result.

Two steps worth checking

  • Evaluating both the function and its derivative at the current approximation
  • Checking what an approximate root means in the original equation

Worked example: improve an approximate root

For real xx, let f(x)=x3+x−5f(x)=x^3+x-5. Find an approximate solution to f(x)=0f(x)=0 by taking two Newton steps from x0=1.5x_0=1.5. Here x0x_0 is the starting guess; x1x_1 and x2x_2 are the next two approximations. Then check a three-decimal answer in the original equation.

Show a hint

First find f′(x)f^{\prime}(x). Each step subtracts the current function value divided by the current slope. Both values must use the same input.

Show the worked solution

  1. 1. Locate the root and choose the update

    We have f(1)=−3f(1)=-3 and f(2)=5f(2)=5. The polynomial is continuous, so a root lies between 1 and 2. Also, f′(x)=3x2+1>0f^{\prime}(x)=3x^2+1>0 for every real xx, so the function is strictly increasing and this root is unique.

    Newton’s method uses the x-intercept of the tangent line as the next guess. Its update is xn+1=xn−f(xn)f′(xn).x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}. The index n=0,1,2,…n=0,1,2,\ldots counts the steps. The denominator must be nonzero. Here it is always positive; for other functions, the method can fail or approach a different root.

  2. 2. Take the first step

    At x0=1.5x_0=1.5, the function value is f(x0)=−0.125f(x_0)=-0.125 and the slope is f′(x0)=7.75f^{\prime}(x_0)=7.75. Therefore x1=1.5−−0.1257.75=4731≈1.516129032.\begin{aligned}x_1&=1.5-\frac{-0.125}{7.75}\\&=\frac{47}{31}\approx1.516129032.\end{aligned} Subtracting a negative correction moves the guess to the right, toward the root.

  3. 3. Recompute both values at the new guess

    Use the unrounded value x1=47/31x_1=47/31 in both parts of the fraction: x2=x1−x13+x1−53x12+1=5094333604≈1.515980240.\begin{aligned}x_2&=x_1-\frac{x_1^3+x_1-5}{3x_1^2+1}\\&=\frac{50943}{33604}\\&\approx1.515980240.\end{aligned} Reusing the old slope 7.757.75 would no longer be the Newton update at x1x_1.

  4. 4. Check the answer’s precision

    Substituting the unrounded x2x_2 gives f(x2)≈1.007×10−7f(x_2)\approx1.007\times10^{-7}. This small remaining function value is called the residual. It checks the equation, but by itself does not generally guarantee a specified number of correct decimal places in x.

    For a separate check, f(1.5159)=−0.000633335321f(1.5159)=-0.000633335321 and f(1.5160)=0.000156096f(1.5160)=0.000156096. Continuity and the sign change place the unique root between these inputs. Every number in that interval rounds to 1.5161.516 to three decimal places. Thus the answer is x≈1.516x\approx1.516.

Your turn: guided step, error check, then a new equation

(1) Guided: keep f(x)=x3+x−5f(x)=x^3+x-5, but start at x0=2x_0=2. Given f(2)=5f(2)=5 and f′(2)=13f^{\prime}(2)=13, find x1x_1. Does the guess move left or right? (2) Diagnose: after the worked example’s first step, a student writes x2=x1−f(x1)/f′(x0)x_2=x_1-f(x_1)/f^{\prime}(x_0). Which input needs changing, and why? (3) Independent: let g(x)=x2−5g(x)=x^2-5. Starting at x0=2x_0=2, calculate two Newton steps and the residual g(x2)g(x_2). Why can you not start this update at x0=0x_0=0?

Show a hint

For (1), subtract 5/135/13 from the starting input. For (2), ask where the new tangent line touches the curve. For (3), derive g′(x)g^{\prime}(x), then evaluate both the numerator and denominator at the latest approximation. Check the denominator before dividing.

Check all three answers

  1. 1. Guided: the positive correction moves the guess left

    x1=2−513=2113≈1.615385.x_1=2-\frac{5}{13}=\frac{21}{13}\approx1.615385. The guess decreases because the function value and slope are both positive. This is one approximation, not an exact solution.

  2. 2. Diagnose: update the slope too

    Replace f′(x0)f^{\prime}(x_0) with f′(x1)f^{\prime}(x_1). The new tangent is taken at x1x_1, so its height and slope must both be evaluated there. The correct step is x2=x1−f(x1)/f′(x1)x_2=x_1-f(x_1)/f^{\prime}(x_1).

  3. 3. Independent: carry out both steps and check the denominator

    Since g′(x)=2xg^{\prime}(x)=2x, the update is xn+1=xn−(xn2−5)/(2xn)x_{n+1}=x_n-(x_n^2-5)/(2x_n), provided xn≠0x_n\ne0. Starting at 2 gives x1=2−−14=94,x2=94−1/169/2=16172≈2.236111.\begin{aligned}x_1&=2-\frac{-1}{4}=\frac94,\\x_2&=\frac94-\frac{1/16}{9/2}\\&=\frac{161}{72}\approx2.236111.\end{aligned}

    The residual is g(x2)=1/5184≈0.000192901g(x_2)=1/5184\approx0.000192901. At x0=0x_0=0, the derivative is zero while g(0)=−5g(0)=-5. The update would divide by zero; the horizontal tangent does not meet the x-axis. Choose a different starting input before applying the method.

If the sign of the correction was unclear, sketch a rising tangent line from a point below the x-axis. If the second step went wrong, put the new input beside both the function value and the derivative before dividing.

What to work through with a tutor

Bring a question, even if you have not started the setup. You can work from the wording together or review a calculation you have tried. Your tutor can help turn an equation into a root-finding problem, check the derivative, choose a starting value from the graph and track where each approximation is used.

For an integration or differential-equation question, your tutor can help identify a suitable method, keep the variables and units consistent, check an antiderivative by differentiation and test a differential-equation solution by substitution. For Fourier series, your tutor can help you identify the period, set up the coefficient integrals and use symmetry to simplify them. Then try a changed question yourself and explain the method before calculating.

Build your review from your current outline

Choose questions from the topics your teacher has assigned. For a numerical method, state the starting value, show the requested steps and check the result in the equation. For integration or differential equations, explain the method and verify your answer. Follow your section’s instructions for calculators, software, notation and assessment requirements.

Independent practice

Choose your next practice step

Course information and further reading

Before requesting a lesson

Can I request tutoring for 201-225-AB?

Yes. Tell us you are taking Mathematical Models 2 at John Abbott, the topic you want to work on and your preferred times. Include the course code if you have it. We confirm tutor availability, lesson details and price before payment.

What should I bring?

Bring your current outline and a question you are allowed to share, if you have them. Any calculation you have started is useful, but you can also ask for help with the first step. For Newton’s method, include any starting value or precision specified in the question; your tutor can help you choose a starting value when none is given.

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