Problem 24 - Calculus Early Transcendentals

Question

(a) If $$G(x)=4 x^{2}-x^{3},\( find$$G^{\prime}(a)\) and use it to find equations of the tangent lines to the curve $$y=4 x^{2}-x^{3}$$ at the points \((2,8)\) and \((3,9)\)

Step-by-Step Solution

Answer

The tangent lines are given by the equations \(y = 4x\) and \(y = -3x + 18\).

Step 1: Differentiate the Function

To find the derivative of the function \(G(x) = 4x^2 - x^3\), we apply the power rule. The differentiation of \(4x^2\) is \(8x\) and the differentiation of \(-x^3\) is \(-3x^2\). So, \(G'(x) = 8x - 3x^2\).

Step 2: Evaluate the Derivative at Given Points

Next, we evaluate the derivative \(G'(x)\) at specific points \(x = 2\) and \(x = 3\) to find \(G'(2)\) and \(G'(3)\). For \(x = 2\), \(G'(2) = 8(2) - 3(2)^2 = 16 - 12 = 4\). For \(x = 3\), \(G'(3) = 8(3) - 3(9) = 24 - 27 = -3\).

Step 3: Equation of the Tangent Line at (2,8)

The equation of a tangent line is in the form \(y - y_1 = m(x - x_1)\) where \(m\) is the slope. At the point \((2,8)\), the slope is \(G'(2) = 4\). Substitute into the formula: \(y - 8 = 4(x - 2)\). Simplifying, we get \(y = 4x\).

Step 4: Equation of the Tangent Line at (3,9)

Using the same tangent line formula, for the point \((3,9)\), with slope \(G'(3) = -3\), we get \(y - 9 = -3(x - 3)\). Simplifying, this becomes \(y = -3x + 18\).

Key Concepts

Tangent Lines

Tangent lines play a crucial role in understanding curves and slopes in calculus. Picture this: when a line just "grazes" or touches a curve at a specific point without crossing it, that's your tangent line. So, why are they important? They show the direction the curve is taking at that precise location. The tangent line to a curve at a given point can be thought of as the curve's "instantaneous direction."
To construct a tangent line, you need two main ingredients:

• The point where the line touches the curve.
• The slope of the curve at that point.

The slope is represented by the derivative of the function at that specific point. Just like the sun casts your shadow in a particular direction, the derivative tells you how steep or flat the shadow (the tangent) should be. This understanding leads us to take a deeper look at derivatives.

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function with respect to a variable. In simpler terms, it's the mathematical tool used to find out how a function behaves as its input changes. Whenever you see a curve and need to know how steep it is at a particular point, you calculate its derivative.
The process of finding a derivative is known as differentiation. At its core, differentiation is about finding how one quantity changes in relation to another. It's like figuring out how fast a car is going by checking how its position changes over time. The derivative is like that speed meter: it tells you how quick the change is happening at every point.
For example, if you have a function like \(G(x) = 4x^2 - x^3\), the derivative \(G'(x)\) gives you a new function that describes the slope of \(G(x)\) at any point. This is where the power rule simplifies our job.

Power Rule

The power rule is a straightforward and powerful rule used in differentiation. It's like a shortcut for finding derivatives of functions that are polynomials. When you want to differentiate a term of the form \(x^n\), the power rule says you bring the exponent down front and reduce the exponent by one. Mathematically, it's expressed as:
\[\frac{d}{dx}[x^n] = nx^{n-1}\]This rule makes it super easy to tackle polynomial functions, sparing you from lengthy calculations with every term.
Let's take the function \(G(x) = 4x^2 - x^3\).

• Apply the power rule to \(4x^2\): multiply by the exponent (2), yielding \(8x^{2-1} = 8x\).
• For \(-x^3\), multiply by the exponent (3), which gives \(-3x^{3-1} = -3x^2\).

After using the power rule, you've swiftly found the derivative, \(G'(x) = 8x - 3x^2\), ready to unveil the slopes needed for tangent line equations. Beautifully simple, isn't it?