/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Find the slope of the function's... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(h(t)=t^{3}, \quad(2,8)\)

Short Answer

Expert verified
The slope at the point is 12, and the tangent line's equation is \(y = 12x - 16\).

Step by step solution

01

Find the derivative of the function

To find the slope of the tangent line to the graph at a specific point, we need to find the function's derivative. The derivative of a function gives us the slope at any point on the graph. Given \(h(t) = t^3\), we differentiate with respect to \(t\): \(h'(t) = 3t^2\).
02

Evaluate the derivative at the given point

Now, we find the slope of the tangent line at the point \((2, 8)\) by evaluating the derivative at \(t = 2\). Substitute \(t = 2\) into the derivative: \(h'(2) = 3(2)^2 = 3(4) = 12\). Thus, the slope at the point \((2, 8)\) is 12.
03

Determine the equation of the tangent line

Having found the slope \(m = 12\), we now use the point-slope form of a line to find the equation of the tangent line. The point-slope formula is \(y - y_1 = m(x - x_1)\). Here, \(m = 12\) and the point is \((x_1, y_1) = (2, 8)\). Substituting these values into the formula gives us: \(y - 8 = 12(x - 2)\).
04

Simplify the equation of the line

Solve this equation for \(y\) to put it in slope-intercept form (\(y = mx + b\)): \(y - 8 = 12x - 24\). Adding 8 to both sides results in \(y = 12x - 16\). This is the equation of the tangent line.

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus. It represents how a function changes as its input changes—essentially, it measures the "rate of change" of a function.
For any function, the derivative at a specific point provides the instantaneous rate of change. This is analogous to finding the slope of the tangent line at that exact point on the graph.
If you picture a curve, the tangent line "touches" the curve only at one point and reflects its immediate direction.
Calculating the derivative involves rules such as the power rule, product rule, quotient rule, and chain rule.
For the function in our exercise, which is given by \( h(t) = t^3 \), the derivative is found using the power rule: \( h'(t) = 3t^2 \). This tells us how steep or flat the function \( h(t) \) is at any point \( t \).
Slope of a Function
The slope of a function at a given point essentially describes how much the function rises or falls as you move along the curve. It's the "tilt" of the function at that point.
The slope is found by evaluating the derivative at a specific point. In this context, the slope of a function corresponds to the slope of its tangent line.
In our example, the derivative \( h'(t) = 3t^2 \) was evaluated at \( t = 2 \) which resulted in \( h'(2) = 12 \). This means the function rises steeply at that point, as compared to other points where the function might appear flatter.
The slope is essentially a real-number value that tells you how rapidly the function is changing at that particular spot.
Point-Slope Form
Once you have the slope of a function at a certain point, you can easily find the equation of a tangent line using the point-slope form formula.
  • This equation is \( y - y_1 = m(x - x_1) \), where \( m \) represents the slope, and \( (x_1, y_1) \) is the specific point the tangent touches the curve.

Using this form allows you to create an equation for the tangent line quickly. For our exercise, with \( m = 12 \) and the point \( (2, 8) \), we applied these in the formula to get \( y - 8 = 12(x - 2) \).
This helped us set up the equation that precisely describes the tangent, reflecting the exact behavior of the curve at that moment.
Differentiation
Differentiation is the process of finding a derivative. It's one of the core techniques in calculus that allows us to determine the rate at which one quantity changes with respect to another.
In practical terms, differentiation gives us a powerful tool to explore how functions behave, making it possible to calculate tangents, velocities, and so much more.
To differentiate a function, you generally apply differentiation rules to transform the function into its derivative. In our exercise with \( h(t) = t^3 \), we differentiated using the power rule, which simply means multiplying by the power and then reducing the power by one.
The result is the derivative \( h'(t) = 3t^2 \), showing the direct relationship between a function and its rate of change. With differentiation, complex problems become manageable, offering insights into curves and trends at any given point.

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Most popular questions from this chapter

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval \(\quad-2

In Exercises \(67-72,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) $$ f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1 $$

Slopes on sine curves a. Find equations for the tangents to the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) at the origin \((m\) a constant \(\neq 0) ?\) Give reasons for your answer. c. For a given \(m,\) what are the largest values the slopes of the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) can ever have? Give reasons for your answer. d. The function \(y=\sin x\) completes one period on the interval \([0,2 \pi],\) the function \(y=\sin 2 x\) completes two periods, the function \(y=\sin (x / 2)\) completes half a period, and so on. Is there any relation between the number of periods \(y=\sin m x\) completes on \([0,2 \pi]\) and the slope of the curve \(y=\sin m x\) at the origin? Give reasons for your answer.

Radians versus degrees: degree mode derivatives What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$ f(h)=\frac{\sin h}{h} $$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$ \lim _{h \rightarrow 0} \frac{\cos h-1}{h} $$ c. Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos \(x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin \(x\) and \(\cos x ?\)

Find the derivatives of the functions in Exercises \(19-40\) $$ f(\theta)=\left(\frac{\sin \theta}{1+\cos \theta}\right)^{2} $$

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