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Chapter 3: Problem 16

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$h(t)=t^{3}+3 t, \quad\quad(1,4)$$

Short Answer

Expert verified
Slope is 6; tangent line equation is \( y = 6t - 2 \).

Step by step solution

01

Differentiate the Function

To find the slope of the tangent line at a specific point, we first need to differentiate the function with respect to \( t \). The given function is \( h(t) = t^3 + 3t \). Differentiating this with respect to \( t \) gives:\[ h'(t) = \frac{d}{dt}(t^3) + \frac{d}{dt}(3t) = 3t^2 + 3. \]
02

Evaluate the Derivative at the Given Point

Next, substitute the \( t \)-coordinate of the given point, \( t = 1 \), into the derivative to find the slope of the tangent line at that point. \[ h'(1) = 3(1)^2 + 3 = 3 + 3 = 6. \]So, the slope of the tangent line at the point (1, 4) is 6.
03

Use the Point-Slope Form to Find the Tangent Line Equation

Now that we have the slope of the tangent line, use the point-slope form of a line equation, which is:\[ y - y_1 = m(x - x_1) \]where \( m \) is the slope, \( (x_1, y_1) \) is the point on the line (1, 4), and we want to find the equation in terms of \( t \) for the variable \( y \). Substituting the known values gives:\[ y - 4 = 6(t - 1). \]
04

Simplify the Equation

To write the equation in its standard form, distribute and simplify:\[ y - 4 = 6(t - 1) \] \[ y - 4 = 6t - 6 \] \[ y = 6t - 6 + 4 \] \[ y = 6t - 2. \]This is the equation of the line tangent to \( h(t) \) at the point (1, 4).

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function grows or shrinks at any given point. It tells us the rate of change of the function's output with respect to changes in the input. Imagine you're on a hill; the derivative is like knowing the steepness or incline at your exact location.
For the function \( h(t) = t^3 + 3t \), taking the derivative means computing the slope at any point \( t \).
  • The process involves applying differentiation rules. For \( t^3 \), we multiply \( 3 \) by \( t^2 \), and for \( 3t \), we simplify it to \( 3 \). This results in the derivative \( h'(t) = 3t^2 + 3 \).

The derivative function now allows us to find the slope of the curve at any specific time \( t \), which is crucial for identifying how quickly the function changes.
Tangent Line
A tangent line lightly "touches" a curve at a specific point and mirrors the curve's immediate direction at that location. Imagine touching a ball with a stick: wherever the stick touches, it's the tangent. This line provides a snapshot of the curve's behavior.
To find the equation of a tangent line at a given point, you need two things: the point itself and the slope of the curve at that point, found using the derivative.
  • Consider the point \( (1, 4) \) on the function \( h(t) \). With the derivative \( h'(t) = 3t^2 + 3 \), we calculate the slope at \( t = 1 \).

This tangent line equation is significant in approximating and interpreting the function's immediate behavior, offering insight into how the function veers near this specific point.
Slope of a Function
The slope of a function is a major concept in calculus and geometry, indicating how steep a line is on a graph. It essentially describes the angle or tilt of a line. For a function defined as \( y = f(x) \), the slope at a particular point is determined by the derivative.
In our example, at \( t = 1 \), the slope is computed by substituting into the derivative \( h'(t) = 3t^2 + 3 \). Plugging \( t = 1 \) gives \( h'(1) = 6 \).
This result means that, at the point (1, 4), the function \( h(t) \) ascends with a steepness of \( 6 \). Understanding the slope helps grasp the rate and direction of change, whether the graph rises or falls as you move along it.
Point-Slope Form
The point-slope form of a line is a practical tool for forming linear equations. It is useful when the slope and a specific point are known. The formula is
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the line.
  • In our example, the slope \( m \) is \( 6 \), and the point is \( (1, 4) \).
Replace these variables in the formula to find the line's equation, \( y - 4 = 6(t - 1) \).
This simple form proves beneficial in highlighting how a line extends from a known point in any graph, connecting the tangent slope with the curve at specified coordinates. Simplifying gives the tangent line's equation in its elementary form \( y = 6t - 2 \).

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Most popular questions from this chapter

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{-x}$$

Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$y=\frac{\sin 2(x+h)-\sin 2 x}{h}$$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

If \(f(x)=x^{n}, n \geq 1,\) show from the definition of the derivative that \(f^{\prime}(0)=0\).

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{\log _{2} t}$$

Find the domain and range of each composite Iunction. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. a. \(y=\sin ^{-1}(\sin x)\) b. \(y=\sin \left(\sin ^{-1} x\right)\)
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