Question

a) Gegeben sei die Kurve \(y=x^{3}-2 x .\) Berechnen Sie die Steigung der Sekante durch die Kurvenpunkte an der Stelle \(x_{1}=1\) und \(x_{2}=\frac{3}{2}\). Vergleichen Sie diese Sekantensteigung mit der Steigung der Tangente an der Stelle \(x_{1}=1\) b) Das Weg-Zeit-Gesetz einer Bewegung sei \(s=3\left[\frac{\mathrm{m}}{\sec ^{2}}\right] t^{2}-8\left[\frac{\mathrm{m}}{\sec }\right] t\) Wie groß ist die Momentangeschwindigkeit zur Zeit \(t=3 \mathrm{sec}\) ? Angabe in \(\mathrm{m} / \mathrm{sec}\). c) Bestimmen Sie jeweils das Differential \(d y\) für die Funktionen \(y=f(x):\) \(\left.\left.\left.c_{1}\right) f(x)=x^{2}+7 x \quad c_{2}\right) f(x)=x^{5}-2 x^{4}+3 \quad c_{3}\right) f(x)=2\left(x^{2}+3\right)\)

Short answer

a) Secant slope: 2.75, Tangent slope at x=1: 1. b) Momentary velocity at t=3: 10 m/sec. c1) dy=(2x+7)dx, c2) dy=(5x^4-8x^3)dx, c3) dy=4xdx.

Step-by-step solution

Calculate the coordinates of the points

First, calculate the coordinates of the points on the curve at the given values of x. For point 1 with x1 = 1, y1 = x1^3 - 2x1 = 1^3 - 2*1 = -1. Thus, Point 1 (x1, y1) = (1, -1). For point 2 with x2 = 3/2, y2 = x2^3 - 2x2 = (3/2)^3 - 2*(3/2) = 27/8 - 3 = 27/8 - 24/8 = 3/8. Thus, Point 2 (x2, y2) = (3/2, 3/8).

Compute the slope of the secant line

The slope of the secant line passing through two points is given by m = (y2 - y1)/(x2 - x1). Substitute the values, m = (3/8 - (-1))/((3/2) - 1) = (3/8 + 8/8)/(3/2 - 1) = 11/8 / 1/2 = 11/4 = 2.75. Thus, the slope of the secant line is 2.75.

Find the derivative of y at x1=1

The slope of the tangent line at a point on the curve is given by the derivative of the function at that point. y = x^3 - 2x Derivative of y (f'(x)) is y' = 3x^2 - 2. At x1=1, y' = 3*(1)^2 - 2 = 3 - 2 = 1. Thus, the slope of the tangent at x=1 is 1.

Compare the slopes

The slope of the secant line through the points (1, -1) and (3/2, 3/8) is 2.75. The slope of the tangent to the curve at point x=1 is 1. Therefore, the secant slope is much steeper than the slope of the tangent.

Momentary velocity

Given the displacement function s = 3[t^2] - 8[t]. The momentary velocity is the derivative of displacement with respect to time (ds/dt). v(t) = d(3t^2 - 8t)/dt = 6t - 8. Thus, v(3) = 6*3 - 8 = 18 - 8 = 10 m/sec.

Compute the differential dy for several functions

For each function, compute the differential dy=f'(x)dx. c1) for f(x) = x^2 + 7x, dy = (2x + 7)dx. c2) for f(x) = x^5 - 2x^4 + 3, dy = (5x^4 - 8x^3)dx. c3) for f(x) = 2(x^2 + 3), dy = 4xdx.

Key concepts

tangent slope calculation

Understanding the slope of the tangent at a given point is a fundamental concept. The slope of the tangent line represents the instantaneous rate of change of the function at that specific point. For a function defined as \( y = x^3 - 2x \), the derivative, which gives us the slope of the tangent, is \( y' = 3x^2 - 2 \).
At the point \( x = 1 \), the value of the derivative is:
\( y'(1) = 3(1)^2 - 2 = 1 \).
So, the slope of the tangent line at \( x = 1 \) is 1.
This calculation tells us how steep the curve is at that point and is vital for understanding the function's behavior around \( x = 1 \).

secant slope comparison

A secant line intersects a curve at two points, and its slope provides a measure of the average rate of change between these points. For the function
\( y = x^3 - 2x \), we are given points \( x_1 = 1 \) and \( x_2 = \frac{3}{2} \). First, find the coordinates:
\( y_1 = 1^3 - 2(1) = -1 \) and \( y_2 = (\frac{3}{2})^3 - 2(\frac{3}{2}) = \frac{3}{8} \).
Using the slope formula:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3/8 + 1}{\frac{3}{2} - 1} = \frac{11/8}{1/2} = 2.75 \).
The secant slope of 2.75 is significantly steeper than the tangent slope of 1 at \( x = 1 \), indicating a rapid increase in the function's value between these points.

momentary velocity determination

The momentary or instantaneous velocity of an object in motion is the first derivative of the position function with respect to time. Given the equation \( s = 3t^2 - 8t \) for displacement, we differentiate to find:
\( v(t) = \frac{ds}{dt} = 6t - 8 \).
To find the velocity at 3 seconds, substitute \( t = 3 \):
\( v(3) = 6(3) - 8 = 10 \text{ m/s} \).
Thus, the object's instantaneous velocity at \( t = 3 \text{ sec} \) is 10 meters per second.

differentiation of functions

Differentiation is a calculus technique used to find the rate at which a function changes at any given point. For each provided function, we compute the differential \( dy \) as follows:

• For \( f(x) = x^2 + 7x \), the differential is \( dy = (2x + 7)dx \).
• For \( f(x) = x^5 - 2x^4 + 3 \), the differential is \( dy = (5x^4 - 8x^3)dx \).
• For \( f(x) = 2(x^2 + 3) \), simplify first to get \( f(x) = 2x^2 + 6 \), then \( dy = 4xdx \).

These differentials help us understand how small changes in \( x \) impact the function \( y \).