Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 3: Problem 59

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=-3 x^{2}+2 ; a=1$$

Short Answer

Expert verified
Answer: The equation of the tangent line to the curve at the point $$a = 1$$ is $$y = -6x + 5$$.

Step by step solution

01

Find the derivative of the function

To find the tangent line's slope at point $$a$$, we need the derivative of the given function. The given function is a quadratic function, and its derivative will give us the function's slope at any point. $$y = -3x^2 + 2$$ $$\frac{dy}{dx} = -6x$$
02

Find the slope of the tangent line at point a

We will now find the slope of the tangent line at point $$a = 1$$ by plugging $$x = 1$$ into the derivative we found in the previous step. $$\frac{dy}{dx} = -6(1)$$ The slope at point $$a$$ is $$-6$$.
03

Find the coordinates of the point a

To find the equation of the tangent line, we also need the $$y$$-coordinate of point $$a$$. Plug $$x = 1$$ into the given function. $$y = -3(1)^2 + 2$$ $$y = -1$$ The point $$a$$ has coordinates $$(1, -1)$$.
04

Find the equation of the tangent line to the curve at point a

Now that we have the slope of the tangent line $$(-6)$$ and the coordinates of point a $$(1, -1)$$, we can use the point-slope formula to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Plug in the coordinates and the slope: $$y - (-1) = -6 (x - 1)$$ Simplify the equation: $$y + 1 = -6x + 6$$ $$y = -6x + 5$$ The equation of the tangent line is $$y = -6x + 5$$.
05

Graph the curve and the tangent line

We are asked to use a graphing utility to graph the curve and the tangent line, so utilize any available graphing software or calculator to plot the given curve $$y = -3x^2 + 2$$ and the tangent line $$y = -6x + 5$$ on the same set of axes. The graph will help to visually validate that the tangent line is, in fact, tangent to the curve at point $$(1, -1)$$.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
Understanding the derivative is crucial when analyzing the behavior of functions. In essence, the derivative of a function represents the rate of change or the slope of the function at a particular point. This mathematical tool allows us to investigate how a function's output value changes as its input value is altered.

A derivative is commonly found using rules of differentiation. For polynomial functions such as the given function in our exercise, \( y = -3x^2 + 2 \), the power rule is applied. Here, we differentiate term by term, bringing down the exponent as a multiplier and reducing the exponent by one. Thus, the derivative \( dy/dx \) corresponds to \( -6x \), which indicates that the slope of the function changes continuously along different points of the curve.
Slope of Tangent
The slope of a tangent line to a curve at a particular point represents the steepness of the curve precisely at that point. In the context of our problem, we find the slope of the tangent line by evaluating the derivative at the specified value of \( x \).

For the function \( y = -3x^2 + 2 \), and its derivative \( -6x \), the slope at the point where \( x = 1 \) is simply the value of the derivative at that point. Plugging \( 1 \) into \( -6x \) gives us \( -6 \) as the slope of the tangent. This value tells us that, at the point where \( x = 1 \), the curve has a downward slope (since it's negative) that is fairly steep, moving 6 units down for every 1 unit across.
Point-Slope Formula
The point-slope formula is an algebraic tool that provides us with a method to find the equation of a line when we know its slope and any point through which it passes. Given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) are the coordinates of the known point, this formula is fundamental in solving our exercise.

By substituting our slope \( -6 \) and point \( (1, -1) \) into the formula, we derive the equation for the tangent line as \( y = -6x + 5 \). This process illustrates how the abstract concept of slope and the concrete coordinates of a particular point come together to define a line precise in both direction and position.
Graphing Functions
Graphing functions and their tangent lines is a powerful way to visualize the concepts of calculus. When we graph the function \( y = -3x^2 + 2 \) and its tangent line at the point \( (1, -1) \), we provide a visual representation of the abstract calculations we have performed.

Using a graphing utility, one can see the parabolic shape of the function and how the tangent line touches the curve only at the one point where \( x = 1 \). This visual aid is not just helpful for verifying our solution but also deepens our understanding of how a tangent line relates to the curve it touches. By observing the graph, students can immediately grasp the slope of the tangent in relation to the curve at any given point and the concept of tangency itself.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The lateral surface area of a cone of radius \(r\) and height \(h\) (the surface area excluding the base) is \(A=\pi r \sqrt{r^{2}+h^{2}}.\) a. Find \(d r / d h\) for a cone with a lateral surface area of \(A=1500 \pi.\) b. Evaluate this derivative when \(r=30\) and \(h=40.\)

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At \(40^{\circ}\) north latitude, the length of a day is approximated by $$D(t)=12-3 \cos \left(\frac{2 \pi(t+10)}{365}\right)$$ where \(D\) is measured in hours and \(0 \leq t \leq 365\) is measured in days, with \(t=0\) corresponding to January 1 a. Approximately how much daylight is there on March 1 \((t=59) ?\) b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March \(1 .\) Convert your answer to units of min/day and explain what this result means. d. Graph the function \(y=D^{\prime}(t)\) using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

a. Derive a formula for the second derivative, \(\frac{d^{2}}{d x^{2}}(f(g(x)))\) b. Use the formula in part (a) to calculate \(\frac{d^{2}}{d x^{2}}\left(\sin \left(3 x^{4}+5 x^{2}+2\right)\right)\)

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free