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Chapter 11: Problem 9

Find the slope of the tangent to the curve at the indicated point. $$\begin{aligned} &y=\sqrt{x} ;(4,2)\\\ &\text { Hint: } x-4=(\sqrt{x}-2)(\sqrt{x}+2) \end{aligned}$$

Short Answer

Expert verified
The slope of the tangent is \( \frac{1}{4} \).

Step by step solution

01

Differentiate the Function

To find the slope of the tangent to the curve, we need to differentiate the function. Given, \( y = \sqrt{x} \), first express it as a power: \( y = x^{1/2} \). The derivative, \( \frac{dy}{dx} \), is \( \frac{1}{2}x^{-1/2} \) which simplifies to \( \frac{1}{2\sqrt{x}} \).
02

Evaluate the Derivative at the Point

We need to find the slope at the point (4, 2). Substitute \( x = 4 \) into the derivative: \( \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4} \).
03

Confirm the Point Lies on the Curve

Verify that (4, 2) is actually a point on the curve \( y = \sqrt{x} \). We substitute \( x = 4 \) into the original function to get \( y = \sqrt{4} = 2 \). The point (4, 2) does lie on the curve.

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Key Concepts

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

Derivative
In calculus, the concept of a derivative is fundamental. It gives us a tool to measure how a function changes as its input changes. In the context of a graph, the derivative at a particular point is the slope of the tangent line to the graph at that point.
When we look at a function, such as \( y = \sqrt{x} \), we're interested in how \( y \) grows as \( x \) increases. To find this rate of change, we differentiate the function. Differentiation tells us the derivative, which is a formula that describes the slope at any point \( x \) on the curve.
  • For \( y = \sqrt{x} \), first express it as a power, \( y = x^{1/2} \).
  • The derivative, \( \frac{dy}{dx} \), measures how \( y \) changes with respect to \( x \), yielding \( \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt{x}} \).
This formula now allows us to find the slope at any specific value of \( x \).
Slope of tangent
The slope of the tangent to a curve at a specific point gives us the direction of the curve at that exact location.
The tangent line just "touches" the curve at a single point without crossing it, representing the instantaneous rate of change.
For practical understanding, think of a straight road that gently kisses the side of a curved path without actually going off into the grass or turning back onto the curved path.
Using our derivative \( \frac{1}{2\sqrt{x}} \), we can substitute the \( x \)-value of the point of interest to find this slope.
  • For the point \((4, 2)\), substitute \( x = 4 \).
  • The slope becomes \( \frac{1}{2\sqrt{4}} = \frac{1}{4} \).
This slope tells you how steep the curve is at \( x = 4 \) and in which direction it's moving.
Differentiation
Differentiation is the process of finding the derivative of a function.
It involves applying rules that simplify the function to its rate of change formula.
In our exercise, this meant converting \( y = \sqrt{x} \) into a simpler power form (like \( x^{1/2} \)) to make differentiation straightforward.
Through applying the power rule,
  • When differentiating, decrease the exponent by one and multiply by the original exponent.
  • For \( x^{1/2} \), the derivative is \( \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).
This shows mathematically how quickly the \( y \)-values (outputs) of the function change as \( x \)-values (inputs) vary, using standard differentiation techniques.
Mathematics education
In mathematics education, teaching calculus concepts like derivatives and slopes of tangents enhances students' problem-solving skills.
These concepts not only apply to academia but also to real-life scenarios, helping students understand the world around them.
In teaching differentiation and slope of tangents:
  • Instructors emphasize understanding the underlying principles, not just mechanical applications.
  • Visual aids, such as graphing software or drawn graphs, show students the real-world applications and representations.
  • Exercises like finding slopes at specific points help build intuitive understanding alongside theoretical knowledge.
Effective education in these areas prepares students for advanced mathematics, physics, engineering, and economics, where these skills become essential.

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Most popular questions from this chapter

By solving the system $$\left\\{\begin{array}{l}y=3 x-1 \\\16 x^{2}-9 y^{2}=144\end{array}\right.$$ show that the line and the hyperbola intersect in exactly one point. Draw a sketch of the situation. This demonstrates that a line that intersects a hyperbola in exactly one point does not have to be a tangent line.

A line is drawn tangent to the ellipse \(x^{2}+3 y^{2}=52\) at the point (2,4) on the ellipse. (a) Find the equation of this tangent line. (b) Find the area of the first-quadrant triangle bounded by the axes and this tangent line.

Graph the equations. $$x^{2}-2 x y+y^{2}+x-y=0$$

Graph the equations. $$(x-y)^{2}=8(y-6)$$

(a) Verify that the point \(P(5,6)\) lies on the hyperbola \(5 y^{2}-4 x^{2}=80\) (b) Find the foci. (c) Compute the lengths of the line segments \(\overline{F_{1} P}\) and \(F_{2} P,\) where \(P\) is the point (5,6) (d) Verify that \(\left|F_{1} P-F_{2} P\right|=2 a\)
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