Problem 55
Question
a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve \(y=x^{3}-4 x+1\) at the point (2,1) b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8 .
Step-by-Step Solution
Verified Answer
a. Equation: \(y = -\frac{1}{8}x + \frac{9}{8}\); b. Smallest slope: -4 at (0,1); c. Tangent equations: \(y = 8x - 15\) and \(y = 8x + 17\).
1Step 1: Find the first derivative
To find the equation of the normal line and other requirements, we first need the derivative of the curve. Given the curve \(y = x^3 - 4x + 1\), the derivative \(\frac{dy}{dx}\) is found using the power rule: \[\frac{dy}{dx} = 3x^2 - 4.\]
2Step 2: Calculate slope at given point
To find the slope of the tangent at the point (2, 1), substitute \(x = 2\) into the derivative: \[\text{slope} = 3(2)^2 - 4 = 12 - 4 = 8.\]
3Step 3: Equation of normal line
The slope of the normal line is the negative reciprocal of the tangent slope. Since the tangent slope at (2,1) is 8, the slope of the normal is \(-\frac{1}{8}\). Using the point-slope form, the equation of the normal line is \[y - 1 = -\frac{1}{8}(x - 2).\] Simplifying gives the equation \(y = -\frac{1}{8}x + \frac{9}{8}\).
4Step 4: Determine smallest slope
The smallest slope of the curve is the minimum value of the derivative \(3x^2 - 4\). Set the derivative of the derivative (second derivative) to zero to find critical points: \[\frac{d^2y}{dx^2} = 6x = 0 \Rightarrow x = 0.\] Substitute \(x = 0\) into the first derivative to get the slope: \[3(0)^2 - 4 = -4.\]
5Step 5: Point with smallest slope
To find the point on the curve with this slope, substitute \(x = 0\) into the original equation: \[y = 0^3 - 4(0) + 1 = 1.\] So, the point is (0,1).
6Step 6: Tangents with specified slope
To find where the slope of the curve is 8, set the derivative equal to 8: \[3x^2 - 4 = 8 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm2.\] Calculate the corresponding \(y\)-values:\(2\): \[y = 2^3 - 4(2) + 1 = 1\] and \(-2\): \[y = (-2)^3 - 4(-2) + 1 = -8 + 8 + 1 = 1.\] The points are (2,1) and (-2,1).
7Step 7: Equations of tangents
Use point-slope form to find tangent equations. For point (2,1): \[y - 1 = 8(x - 2) \Rightarrow y = 8x - 15.\] For point (-2,1): \[y - 1 = 8(x + 2) \Rightarrow y = 8x + 17.\]
Key Concepts
Tangent SlopeFirst DerivativeCritical PointsEquation of Normal Line
Tangent Slope
The tangent slope of a curve at a particular point is crucial in understanding how steep the curve is at that exact location. It's essentially the slope of the 'tangent line,' which just touches the curve without crossing it at the point of tangency. To figure out the tangent slope, we need to first find the derivative of the curve's equation, as it gives us a formula for the slope of the tangent line at any point along the curve.
Once we have the derivative, it's a simple matter of substituting the x-coordinate of the point of interest into the derivative to compute the slope at that point. For example, if the derivative is given as \(3x^2 - 4\) and we need the slope at \(x = 2\), we plug in \(x = 2\) to get \(3(2)^2 - 4 = 8\). This tells us the tangent slope at \(x = 2\) is 8.
Understanding tangent slopes helps in many practical applications, such as determining instantaneous rates of change in real-world contexts like physics or economics.
Once we have the derivative, it's a simple matter of substituting the x-coordinate of the point of interest into the derivative to compute the slope at that point. For example, if the derivative is given as \(3x^2 - 4\) and we need the slope at \(x = 2\), we plug in \(x = 2\) to get \(3(2)^2 - 4 = 8\). This tells us the tangent slope at \(x = 2\) is 8.
Understanding tangent slopes helps in many practical applications, such as determining instantaneous rates of change in real-world contexts like physics or economics.
First Derivative
The first derivative of a function is a powerful mathematical tool that mirrors the slope of the curve itself. Mathematically, the first derivative \(\frac{dy}{dx}\) is derived using differentiation techniques. For the specific curve \(y = x^3 - 4x + 1\), using the power rule gives us the first derivative \(3x^2 - 4\). This is a function that we can use to find the slope of the tangent line at any point x.
The first derivative is also essential in calculus as it helps us identify various properties of the curve such as increasing and decreasing regions, and assists in finding critical points which are used to identify minimum and maximum points.
By evaluating this derivative at different x values, one can observe how the steepness of the curve changes. This insight is vital, especially when analyzing phenomena that have non-constant rates of change.
The first derivative is also essential in calculus as it helps us identify various properties of the curve such as increasing and decreasing regions, and assists in finding critical points which are used to identify minimum and maximum points.
By evaluating this derivative at different x values, one can observe how the steepness of the curve changes. This insight is vital, especially when analyzing phenomena that have non-constant rates of change.
Critical Points
Critical points on a curve occur where the derivative is zero or undefined. These are points where the curve has the possibility to change direction, making them potential locations for local maxima or minima - commonly referred to as turning points. To find critical points, you need to set the first derivative equal to zero and solve for x.
In our example, with the derivative function \(3x^2 - 4\), solving \(3x^2 - 4 = 0\) tells us that \(x = 0\). These critical points are then substituted back into the original function to find the corresponding y-values. This helps in pinpointing the coordinates on the curve where slopes change from positive to negative or vice versa.
Finding the critical points is a fundamental step in the process of graph sketching and optimization problems, where determining the highest or lowest points of a function is necessary.
In our example, with the derivative function \(3x^2 - 4\), solving \(3x^2 - 4 = 0\) tells us that \(x = 0\). These critical points are then substituted back into the original function to find the corresponding y-values. This helps in pinpointing the coordinates on the curve where slopes change from positive to negative or vice versa.
Finding the critical points is a fundamental step in the process of graph sketching and optimization problems, where determining the highest or lowest points of a function is necessary.
Equation of Normal Line
The equation of a normal line to a curve at a given point is derived from its relationship to the tangent line. Specifically, the normal line is perpendicular to the tangent line at the point of tangency. Hence, its slope is the negative reciprocal of the tangent slope.
If the tangent slope at a point is given as 8, then the normal slope would be \(-\frac{1}{8}\). Using the point-slope formula for a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, \((x_1, y_1)\) is the point, we can find the equation of the normal line. For instance, if our point is (2,1), the normal line's equation becomes \(y - 1 = -\frac{1}{8}(x - 2)\), simplifying to \(y = -\frac{1}{8}x + \frac{9}{8}\).
Understanding how to derive the equation of the normal line is useful in many aspects of geometry and physics, where perpendicular lines play a key role in constructing precise models and solving problems related to perpendicular vectors and forces.
If the tangent slope at a point is given as 8, then the normal slope would be \(-\frac{1}{8}\). Using the point-slope formula for a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, \((x_1, y_1)\) is the point, we can find the equation of the normal line. For instance, if our point is (2,1), the normal line's equation becomes \(y - 1 = -\frac{1}{8}(x - 2)\), simplifying to \(y = -\frac{1}{8}x + \frac{9}{8}\).
Understanding how to derive the equation of the normal line is useful in many aspects of geometry and physics, where perpendicular lines play a key role in constructing precise models and solving problems related to perpendicular vectors and forces.
Other exercises in this chapter
Problem 55
How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. Derivative of arcsine Assume that \(y=
View solution Problem 55
The equations give the position \(s=f(t)\) of a body moving on a coordinate line ( \(s\) in meters, \(t\) in seconds). Find the body's velocity, speed, accelera
View solution Problem 55
Does knowing that a function \(f(x)\) is differentiable at \(x=x_{0}\) tell you anything about the differentiability of the function \(-f\) at \(x=x_{0} ?\) Giv
View solution Problem 56
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(3 \theta e^{-\theta}\right)$$
View solution