Question

Tangent line is \(p_{1}\) Let \(f\) be differentiable at \(x=a\). a. Find the equation of the line tangent to the curve \(y=f(x)\) at \((a, f(a))\). b. Verify that the Taylor polynomial \(p_{1}\) centered at \(a\) describes the tangent line found in part (a).

Short answer

Question: Find the equation of a tangent line to a curve at a given point and verify that the first-order Taylor polynomial describes this tangent line. Answer: To find the equation of a tangent line, calculate the function's derivative at the given point (f'(a)) and use the point-slope formula: y - f(a) = f'(a)(x - a). Next, find the first-order Taylor polynomial using the simplified Taylor polynomial formula: p_1(x) = f(a) + f'(a)(x - a). Lastly, compare the tangent line equation and the first-order Taylor polynomial. If they match, it verifies that the Taylor polynomial accurately describes the tangent line.

Step-by-step solution

Find the equation of the tangent line

To find the equation of the tangent line, we will need to determine the slope of the line at the given point \((a, f(a))\). Since \(f\) is differentiable at \(x=a\), we can calculate the derivative \(f'(a)\), which represents the instantaneous rate of change (or the slope) of the tangent line at this point. The equation of the tangent line is given by the point-slope formula, which is: \(y - f(a) = f'(a) (x - a)\)

Find the first-order Taylor polynomial

The Taylor polynomial of a function centered at \(x=a\) is given by the formula: \(p_{n}(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \cdots + \frac{1}{n!}f^{(n)}(a)(x-a)^n\) Since we are interested in the first-order Taylor polynomial (\(p_1\)), we can ignore the terms with higher-order derivatives and simplify the expression to: \(p_{1}(x) = f(a) + f'(a)(x - a)\)

Verify that the tangent line and the Taylor polynomial match

We now compare the equation of the tangent line and the first-order Taylor polynomial. From Steps 1 and 2, we have: Tangent Line: \(y - f(a) = f'(a)(x - a)\) First-Order Taylor Polynomial: \(p_{1}(x) = f(a) + f'(a)(x - a)\) We can rewrite the tangent line equation as: \(y = f(a) + f'(a)(x - a)\) Now we can see that the equations for the tangent line and the first-order Taylor polynomial are indeed the same. Therefore, the Taylor polynomial \(p_1\) centered at \(a\) describes the tangent line found in part (a).

Key concepts

Differentiable Functions

When we study calculus, understanding the meaning of 'differentiable functions' becomes critical. A function is said to be differentiable at a point if it has a defined derivative there; in simpler terms, you can smoothly draw a tangent line at that point without lifting your pen.

Consider a curve representing the function on a graph. At each point on the curve, if you can imagine a straight line that just 'kisses' the curve, that's a tangent line, representing the instantaneous rate of change of the function at that point. Differentiable functions allow us to use these tangent lines to approximate function values near our point of interest.

Another important characteristic is that differentiable functions must be continuous. A sudden jump or break would mean you can't draw that smooth tangent line. Be mindful, however that being continuous doesn't always mean the function is differentiable. Think of a sharp point, like the tip of a 'V': It's continuous, but there's no single tangent line you can draw at the tip.

Taylor Polynomial

Let's talk about the Taylor polynomial, a powerful tool in calculus for approximating the values of functions. Named after mathematician Brook Taylor, the Taylor polynomial provides an approximate representation of a function around a specific point, usually denoted as 'a'. This is based on the concept that we can use a polynomial — a sum of power functions like quadratics, cubics, and so on — to mimic the behavior of a more complex function near 'a'.

The Power of Polynomials

What makes a Taylor polynomial so special is its ability to approximate functions to a very high accuracy using just algebraic operations. The formula combines the function's value and its derivatives at 'a', each multiplied by powers of \(x-a\). The 'n' in the polynomial's subscript refers to the degree - the number of terms used. A first-order Taylor polynomial incorporates just the function's value and its first derivative, making it a straight line, which, you've guessed it, represents the tangent line at the point 'a'.

Point-Slope Formula

Moving to the very heart of finding equations of lines in coordinate geometry: the point-slope formula. It's based on a simple premise — if you know a single point on a line and the line’s slope, you've got enough information to write the line's equation. The generic formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \(\left(x_1, y_1\right)\) is the known point.

Essential for calculus students, the point-slope formula shines when dealing with curves. Once the slope of the tangent line (which is the derivative of the function at that point) is found, and given a specific point on the function, we apply this formula. It’s especially useful because it doesn't require the line to be in any special position, such as passing through the origin or being parallel to an axis. Through this lens, calculus and algebra intermingle, beautifully bridging the gap between graphical representations and algebraic expressions.

Instantaneous Rate of Change

Finally, let's delve into the 'instantaneous rate of change'. Imagine you're driving and your friend asks, 'How fast are you going right now?' Not 'on average since we left home', but right at this instant. In calculus, that's essentially what the derivative of a function at a point gives us — the instantaneous rate of change.

It's what a tangent line's slope represents: the rate at which the function's value is changing at that precise point. In economic terms, it's like looking at the speed of price changes for a stock; in physics, it's akin to calculating the exact velocity of an object at a single point in time.

Connecting Concepts

When we bring differentiable functions, Taylor polynomials, and the point-slope formula together, we get a complete picture. The derivative gives us the slope for the point-slope formula, which in turn helps us write the equation for the tangent line — and that tangent line's equation is the same as that of the first-order Taylor polynomial. It's a round of conceptual relay race where each concept passes the baton to the next, resulting in a beautiful synthesis of ideas.