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Chapter 3: Problem 38

angent lines a. Find the derivative function \(f^{\prime}\) for the following functions \(f\) b. Find an equation of the line tangent to the graph of \(f\) at \((a, f(a))\) for the given value of \(a\). $$f(x)=\sqrt{x+2} ; a=7$$

Short Answer

Expert verified
Answer: The equation of the tangent line is \(y = \frac{1}{6}(x - 7) + 3\).

Step by step solution

01

Find the derivative of the function \(f(x)\)

We have \(f(x) = \sqrt{x+2}\). The function is in the form of \(g(x) = \sqrt{h(x)}\) where \(g(x) = \sqrt{x}\) and \(h(x) = x + 2\). We can use the chain rule to find the derivative of \(f(x)\). The chain rule states that \((g(h(x)))'= g'(h(x)) \cdot h'(x)\). We will find the derivatives of \(g(x)\) and \(h(x)\) first. $$g'(x) = \frac{d\sqrt{x}}{dx} = \frac{1}{2\sqrt{x}}$$ $$h'(x) = \frac{d(x+2)}{dx} = 1$$ Now, we will apply the chain rule to find the derivative of \(f(x)\): $$f^{\prime}(x) = g'(h(x)) \cdot h'(x) = \frac{1}{2\sqrt{h(x)}} \cdot 1 = \frac{1}{2\sqrt{x+2}}$$
02

Evaluate the derivative at the point \(a=7\)

We found the derivative \(f^{\prime}(x) = \frac{1}{2\sqrt{x+2}}\). Now we will evaluate it at \(a=7\) to find the slope of the tangent line at the given point. $$f^{\prime}(7) = \frac{1}{2\sqrt{7+2}} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \cdot 3} = \frac{1}{6}$$
03

Find the slope of the tangent line

The slope of the tangent line is the value of the derivative at the given point. We found that value to be \(\frac{1}{6}\).
04

Calculate the equation of the tangent line

To find the equation of the tangent line, we will use the point-slope formula: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the tangent point and \(m\) is the slope of the tangent line. In this case, we have the following values: - The point \((x_1, y_1)\) is \((a, f(a)) = (7, f(7)) = (7, \sqrt{7+2}) = (7, \sqrt{9}) = (7, 3)\). - The slope \(m = \frac{1}{6}\). Now, we will plug these values into the point-slope formula: $$y - 3 = \frac{1}{6}(x - 7)$$ Now we can rearrange the equation to get the final equation of the tangent line: $$y = \frac{1}{6}(x - 7) + 3$$

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Key Concepts

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

Derivative Function
In calculus, the derivative function represents the rate at which a function's output changes as its input changes. Essentially, it gives us the slope of the tangent line to the curve at any point. The process of finding the derivative is called differentiation. For a function f(x), the derivative is denoted by f'(x) or \frac{df}{dx}.

For example, to find the derivative of f(x) = \(\sqrt{x+2}\), we differentiate the function with respect to x. The resulting derivative function gives us the slope of the tangent line at any point along f(x), which is crucial for solving many problems in calculus, such as finding maximum and minimum values or solving related rates problems.
Chain Rule
The chain rule is a fundamental rule in calculus used for differentiating compositions of functions. It is based on the principle that if you have two functions鈥攕ay, g(x) and h(x)鈥攚here one function is nested inside of the other, like g(h(x)), the derivative of this composition is the product of the derivative of the outside function evaluated at the inside function and the derivative of the inside function.

When we find g'(h(x)) \cdot h'(x), we're applying the chain rule. This is precisely what was done in the step-by-step solution to differentiate f(x) = \(\sqrt{x+2}\). Understanding the chain rule is essential for working with more complex functions where multiple functions are composed together.
Point-Slope Formula
The point-slope formula is a method for finding the equation of a line when you know a point on the line and its slope. The general form of the point-slope formula is y - y_1 = m(x - x_1), where (x_1, y_1) is the known point and m is the slope of the line.

In the original exercise, we used this formula to write the equation of the tangent line to the graph of f(x) at a specific point. This line's slope is the same as the derivative of f(x) at the given point, and using the coordinates of that point along with the point-slope formula makes it straightforward to write down the equation of the tangent line.
Evaluating Derivatives
The process of evaluating derivatives involves calculating the derivative of a function at a given point. This is a vital step when finding the slope of the tangent line to a function's graph at a particular point鈥攖he derivative tells us the slope, and when we evaluate it at a specific x-value, we get the slope at that point.

In the solution provided, we first found the derivative of f(x) = \(\sqrt{x+2}\) using the chain rule. We then evaluated this derivative function at x = 7 to get the slope of the tangent line at (7, f(7)). Evaluating derivatives like this allows us to answer many practical questions about how functions change and is essential in optimization problems and in understanding the behavior of functions.

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Most popular questions from this chapter

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at \(9.00 ?\) (Hint: Use the Law of Cosines.)

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$x^{2 / 3}+y^{2 / 3}=2 ;(1,1)$$

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x^{3}=\frac{x+y}{x

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?
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