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 3.11.14a

Chapter 3, Problem 3.11.14a

Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of 4 m/s.
a. At what rate is the area of the triangle changing when the legs are 5 m long?

Verified step by step guidance

Identify the relationship between the hypotenuse and the legs of the isosceles right triangle. In an isosceles right triangle, the hypotenuse \( c \) is related to the legs \( a \) by the equation \( c = a\sqrt{2} \).

Differentiate the equation \( c = a\sqrt{2} \) with respect to time \( t \) to find the relationship between the rates of change of the hypotenuse and the legs. This gives \( \frac{dc}{dt} = \sqrt{2} \frac{da}{dt} \).

Substitute the given rate of change of the hypotenuse \( \frac{dc}{dt} = -4 \) m/s into the differentiated equation to solve for \( \frac{da}{dt} \), the rate of change of the legs.

Use the formula for the area \( A \) of an isosceles right triangle, \( A = \frac{1}{2}a^2 \), and differentiate it with respect to time \( t \) to find \( \frac{dA}{dt} \). This gives \( \frac{dA}{dt} = a \frac{da}{dt} \).

Substitute \( a = 5 \) m and the previously calculated \( \frac{da}{dt} \) into the differentiated area formula to find the rate at which the area is changing, \( \frac{dA}{dt} \).

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Key Concepts

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

Related Rates

Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to relate the rate of change of the hypotenuse to the rate of change of the area of the triangle. This requires the use of derivatives to express how changes in one variable affect another over time.

Area of a Triangle

The area of a triangle can be calculated using the formula A = (1/2) * base * height. For an isosceles right triangle, the legs are equal, and both serve as the base and height. Understanding how to express the area in terms of the leg length is crucial for determining how the area changes as the triangle shrinks.

Pythagorean Theorem

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is essential for relating the lengths of the legs of the triangle to the hypotenuse, allowing us to express the leg lengths in terms of the hypotenuse as it changes over time.

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a. Verify that the given point lies on the curve.

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a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.

f(w) = w³ -w / w

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a. Express the angle of elevation θ from the biologist to the falcon as a function of the height h of the bird above the ground. (Hint: The vertical distance between the top of the cliff and the falcon is 80−h.)

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Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>

a. Calculate the average velocity of the airliner during the first 1.5 hours of the trip (0 ≤ t ≤ 1.5).

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Textbook Question

{Use of Tech} Approximating derivatives Assuming the limit exists, the definition of the derivative f′(a) = lim h→0 f(a + h) − f(a) / h implies that if ℎ is small, then an approximation to f′(a) is given by

f' (a) ≈ f(a+h) - f(a) / h. If ℎ > 0 , then this approximation is called a forward difference quotient; if ℎ < 0 , it is a backward difference quotient. As shown in the following exercises, these formulas are used to approximate f′ at a point when f is a complicated function or when f is represented by a set of data points. <IMAGE>

Let f (x) = √x.

a. Find the exact value of f' (4).

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Textbook Question

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a. Use implicit differentiation to find dy/dx.

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