Understanding Slopes and Their Applications

This problem requires us to calculate various types of slopes based on given functions and points and then match them to the corresponding values provided in List II. We will analyze each item in List I individually.

Let's break down each part of the question:

A. Slope of the tangent to curve \( x^3 - 2x \) at \( x = 2 \)

The slope of the tangent to a curve \( y = f(x) \) at a specific point \( x \) is given by the derivative of the function at that point, i.e., \( m_t = f'(x) \).

Given the curve \( y = x^3 - 2x \).

We need to find the derivative of \( y \) with respect to \( x \):

\( \frac{dy}{dx} = \frac{d}{dx}(x^3 - 2x) \)

\( \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) \)

Using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) and the constant multiple rule:

\( \frac{dy}{dx} = 3x^{3-1} - 2 \cdot 1 \cdot x^{1-1} \)

\( \frac{dy}{dx} = 3x^2 - 2 \)

Now, we need to evaluate this derivative at \( x = 2 \) to find the slope of the tangent at that point:

Slope \( m_t = \left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 2 \)

\( m_t = 3(4) - 2 \)

\( m_t = 12 - 2 \)

\( m_t = 10 \)

So, the slope of the tangent to the curve \( x^3 - 2x \) at \( x = 2 \) is 10. This matches option II in List II.

B. Slope of line passing through the points (0,2) and (5,-6)

The slope of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:

\( m = \frac{y_2 - y_1}{x_2 - x_1} \)

Given the points (0,2) and (5,-6). Let \( (x_1, y_1) = (0,2) \) and \( (x_2, y_2) = (5,-6) \).

Using the formula:

\( m = \frac{-6 - 2}{5 - 0} \)

\( m = \frac{-8}{5} \)

So, the slope of the line passing through the points (0,2) and (5,-6) is \( -\frac{8}{5} \). This matches option III in List II.

C. Point at which the tangent to the curve \( y = \sqrt{4x - 3} \) has its slope \( \frac{2}{3} \)

First, we need to find the general formula for the slope of the tangent to the curve \( y = \sqrt{4x - 3} \). This requires finding the derivative \( \frac{dy}{dx} \).

Rewrite the function as \( y = (4x - 3)^{1/2} \).

Using the chain rule \( \frac{dy}{dx} = \frac{d}{du}(u^{1/2}) \cdot \frac{du}{dx} \) where \( u = 4x - 3 \):

\( \frac{du}{dx} = \frac{d}{dx}(4x - 3) = 4 \)

\( \frac{dy}{du} = \frac{d}{du}(u^{1/2}) = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \)

So, \( \frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} \cdot 4 \)

\( \frac{dy}{dx} = \frac{4}{2\sqrt{4x - 3}} \)

\( \frac{dy}{dx} = \frac{2}{\sqrt{4x - 3}} \)

We are given that the slope of the tangent is \( \frac{2}{3} \). So, we set the derivative equal to \( \frac{2}{3} \):

\( \frac{2}{\sqrt{4x - 3}} = \frac{2}{3} \)

To solve for \( x \), we can cross-multiply or notice that the numerators are equal, so the denominators must be equal:

\( \sqrt{4x - 3} = 3 \)

Square both sides to eliminate the square root:

\( (\sqrt{4x - 3})^2 = 3^2 \)

\( 4x - 3 = 9 \)

\( 4x = 9 + 3 \)

\( 4x = 12 \)

\( x = \frac{12}{4} \)

\( x = 3 \)

Now that we have the x-coordinate of the point, we need to find the corresponding y-coordinate by substituting \( x = 3 \) back into the original curve equation \( y = \sqrt{4x - 3} \):

\( y = \sqrt{4(3) - 3} \)

\( y = \sqrt{12 - 3} \)

\( y = \sqrt{9} \)

\( y = 3 \) (Assuming the positive root for the curve definition)

The point at which the tangent to the curve \( y = \sqrt{4x - 3} \) has its slope \( \frac{2}{3} \) is (3,3). This matches option IV in List II.

D. Slope of normal to the curve \( y = \frac{x-2}{x-1} \) at \( x = 10 \)

First, we need to find the slope of the tangent to the curve \( y = \frac{x-2}{x-1} \) at \( x = 10 \). This requires finding the derivative \( \frac{dy}{dx} \).

Using the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \), where \( u = x-2 \) and \( v = x-1 \):

\( \frac{du}{dx} = \frac{d}{dx}(x-2) = 1 \)

\( \frac{dv}{dx} = \frac{d}{dx}(x-1) = 1 \)

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

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

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

Now, evaluate the derivative at \( x = 10 \) to find the slope of the tangent \( m_t \) at this point:

\( m_t = \left. \frac{dy}{dx} \right|_{x=10} = \frac{1}{(10-1)^2} \)

\( m_t = \frac{1}{(9)^2} \)

\( m_t = \frac{1}{81} \)

The slope of the normal \( m_n \) to the curve at a point is the negative reciprocal of the slope of the tangent \( m_t \) at that point, provided \( m_t \neq 0 \). The relationship is \( m_n = -\frac{1}{m_t} \).

Using the calculated slope of the tangent \( m_t = \frac{1}{81} \):

\( m_n = -\frac{1}{1/81} \)

\( m_n = -81 \)

So, the slope of the normal to the curve \( y = \frac{x-2}{x-1} \) at \( x = 10 \) is -81. This matches option I in List II.

Summary of Matches

Based on our calculations, the matches are:

• A: Slope of the tangent to \( x^3 - 2x \) at \( x = 2 \) is 10 (II).
• B: Slope of line through (0,2) and (5,-6) is \( -\frac{8}{5} \) (III).
• C: Point where tangent to \( y = \sqrt{4x - 3} \) has slope \( \frac{2}{3} \) is (3,3) (IV).
• D: Slope of normal to \( y = \frac{x-2}{x-1} \) at \( x = 10 \) is -81 (I).

This gives us the matching A-II, B-III, C-IV, D-I.

Revision Table: Calculus and Coordinate Geometry Slopes

Additional Information on Slopes and Derivatives

The concept of slope is fundamental in both geometry and calculus. In geometry, it describes the inclination of a straight line. In calculus, the derivative extends this idea to curves, allowing us to find the slope of the tangent line at any point on the curve. This tangent slope represents the instantaneous rate of change of the function.

Derivative Rules Used:

• Power Rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \)
• Constant Multiple Rule: \( \frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x)) \)
• Sum/Difference Rule: \( \frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x)) \)
• Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) (used for \( \sqrt{4x-3} \))
• Quotient Rule: \( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \) (used for \( \frac{x-2}{x-1} \))

The slope of the normal is important because it represents the direction perpendicular to the curve at a given point. Normals are used in various applications, such as determining the direction of forces or the path of light reflecting off a surface.