Calculus Online: Lab 6

Welcome to Lab 6 of Math 100 Sections 103, 104, 107 and 109.


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  3. All submissions must be received by 11:59 pm on 4 December 1998.


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More Approximations

In the last lab, we saw how we could use the derivative to generate approximate solutions to two different problems. Newton's Method helped us to find approximate solutions to algebraic equations while Euler's Method generated approximate solutions to differential equations. Underlying both of these methods was the fundamental fact about the graph of a function:

Near the point x0, the tangent line at x0 provides a good approximation for the graph.
In this lab, we'll explore this idea further and expand on it.

Question 1: The Linear Approximation

Newton's Method and Euler's Method worked by approximating a general graph with the tangent line at some point. Of course, we can do this with any function, and we call it a linear approximation since we approximating the function with a linear function.

In this question, we'll build the linear approximation for the function

y(x) = \sqrt{x} 

at the point x0 = 1.

Part a (2 marks)

Select "Question 1a" above and enter the value of the function $  y(x) = \sqrt{x}  $ at x0 = 1. Then either press "Return" or click the "OK" button. You will see the value you have entered appear on the graph.

Part b (2 marks)

Select "Question 1b" above and enter the value of the derivative of the function $  y(x) = \sqrt{x}  $ at x0 = 1. When you press "Return" or click the "OK" button, you will see the tangent line appear.

This is what we are after: whereas the original function $ 
y(x) = \sqrt{x}  $ can be difficult to understand thoroughly, the tangent line is very easy to work with.

Part c (2 marks)

Select "Question 1c" and enter the value given by the linear approximation to the function at x = 2. You will see your result displayed together with an indication of how far the approximation differs from the true value $  \sqrt{2} = 
1.414.... $

It's easy to find $  \sqrt{2}  $ with a calculator. But here, you have generated a reasonable approximation using calculations you could probably do in your head. In the next questions, we'll see how to improve our approximation with just a little more work.

Question 2: Quadratic Approximations

In the previous question, we approximated a graph by a straight line. You can see that as we travel further away from x0, the point at which we found the tangent line, the approximation becomes worse.

But why does the approximation get worse? Because the graph is curved and we are approximating it by a straight line. If we instead approximated the graph by something which was curved, then perhaps we would find a better approximation.

In this question, we will approximate a function f(x) by a quadratic function

p(x) = a + b x + c x2

whose graph is a parabola. By considering the curvature of the graph in this way, you can see that the approximation becomes better over a larger range.

Part a (2 marks)

Select "Question 2a" above and position the constant coefficient a to give the correct value for the function at x0 = 0. Another way to think of this is that you are approximating the function by a constant function in the vicinity of x0 = 0.

Part b (2 marks)

Select "Question 2b" and position the coefficient of the linear term to give the best approximation to the graph at x0 = 0. This is, in fact, the linear approximation--the kind of approximation we considered in Question 1. The graph you have constructed so far should be the tangent line at x0 = 0. Notice that the coefficient of x is the derivative of the function at x0 = 0.

Part c (2 marks)

Select "Question 2c" and position the coefficient of the quadratic term to give the best approximation to the graph at x0 = 0. Notice that adding in the quadratic term gives a better approximation over a wider range than the linear approximation.

Part d (2 marks)

Let's think about what the coefficients of the approximation mean. The constant coefficient is the value of the function at 0 --that is, p(0) = f(0). In the same way, the coefficient of x is f'(0), the derivative at 0. This means that p'(0) = f'(0). This produces the tangent line for the linear approximation.

What should the coefficient of x2 be? Well, if we want the parabola to be as curved as the original graph near x = 0, we should make p''(0) = f''(0) since the second derivative controls how a graph curves.

To be sure you understand this, enter the second derivative of the function f(x) at x = 0 in the window below. Be careful, it is not just the coefficient c that you found above, but it is closely related to it. (What is the derivative of the quadratic at x = 0? Differentiate the polynomial p(x) = a + b x + c x2 twice.)

Question 3: Taylor Polynomials (8 marks)

In the last two problems, we have found polynomials which approximate a given function by equating a certain number of derivatives of the function and the polynomial at some point. For instance, in Question 1, we formed the linear approximation by equating the value of the function and its first derivative to those same quantities in a linear function. Then in Question 2, we formed a quadratic approximation by equating the value of a function, its derivative and its second derivative with the corresponding quantities in a quadratic polynomial.

Of course, there is no reason to stop there. In this question, we will form a much better approximation from a higher degree polynomial in the same way. We will consider the function f(x) = ex . Notice that f(0) = 1 and also f'(0) = 1 . In fact, no matter how many times we differentiate f(x) = ex, the value is still 1 at x = 0.

We will create a polynomial

p(x) = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6 + hx^7 

such that p(0) = f(0) and in addition, the first seven derivatives of p and f agree at x = 0. Enter the coefficients in the boxes below and press "Return" after you make each entry. In this way, you can build the approximation step by step. You should notice that it tracks along the graph of f(x) = ex extremely well as you enter the coefficients. Also displayed for you is the value p(1). This value is approximating $  f(1) = e \approx 2.71828... $

The windows in which you are to enter the coefficients can perform division for you: for instance, the window will understand what you mean if you enter "1/2". In fact, you may enter "1/2/4" if you mean 1/8.

Approximating functions in this way is a nice thing to do: it is very difficult to compute the values of the function f(x) = ex from scratch. Here we have produced a polynomial, in a simple way, which does a very good job of approximating the function. In essence, we have replaced a complicated function with a simple one. We can produce even better approximations by taking a higher order polynomial. Such polynomials are called Taylor polynomials. Next term we'll see some more uses of this kind of approximation.

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We hoped you liked the labs this term. Good luck on the final and we'll see you again next term.