Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.40b

Chapter 3, Problem 3.40b

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = √3x; a= 12

Verified step by step guidance

Step 1: Identify the function and the point of tangency. The function given is $ f(x) = \sqrt{3x} $ and the point of tangency is $ (a, f(a)) $ where $ a = 12 $.

Step 2: Calculate $ f(a) $. Substitute $ a = 12 $ into the function to find $ f(12) = \sqrt{3 \times 12} $.

Step 3: Find the derivative of the function $ f(x) = \sqrt{3x} $. Use the chain rule: $ f'(x) = \frac{d}{dx}(3x)^{1/2} = \frac{1}{2}(3x)^{-1/2} \cdot 3 $.

Step 4: Evaluate the derivative at $ x = 12 $. Substitute $ x = 12 $ into $ f'(x) $ to find $ f'(12) $.

Step 5: Use the point-slope form of a line to write the equation of the tangent line. The formula is $ y - f(a) = f'(a)(x - a) $. Substitute $ f(12) $ and $ f'(12) $ into this equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function at a point measures the rate at which the function's value changes as its input changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the graph of the function at that point.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. The equation of the tangent line can be expressed in point-slope form, which is derived from the slope and the coordinates of the point of tangency.

Point-Slope Form

The point-slope form of a linear equation is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope (derivative) and the point of tangency are known. By substituting the appropriate values, one can easily derive the equation of the tangent line.

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Textbook Question

A race Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions θ(t) and ϕ(t), respectively, where 0≤t≤4 and t is measured in minutes (see figure). These angles are measured in radians, where θ=ϕ=0 represent the starting position and θ=ϕ=2π represent the finish position. The angular velocities of the runners are θ′(t) and ϕ′(t). <IMAGE>

b. Which runner has the greater average angular velocity?

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Textbook Question

Derivatives using tables Let $h(x)=f(g(x))$ and $p(x)=g(f(x))$ . Use the table to compute the following derivatives.

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b. $h prime of 2$

Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

x = e^y; (2, ln 2)

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Textbook Question

{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>

b. Find dx/dt and interpret the meaning of this derivative.

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(x) = 1/x+1; a = -1/2;5

Textbook Question

79–82. {Use of Tech} Visualizing tangent and normal lines

b. Graph the tangent and normal lines on the given graph.

(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

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