/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Area The area \(A\) of a triangl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 19

Area The area \(A\) of a triangle with sides of lengths \(a\) and \(b\) enclosing an angle of measure \(\theta\) is $$A=\frac{1}{2} a b \sin \theta.$$ a. How is \(d A / d t\) related to \(d \theta / d t\) if \(a\) and \(b\) are constant? b. How is \(d A / d t\) related to \(d \theta / d t\) and \(d a / d t\) if only \(b\) is constant? c. How is \(d A / d t\) related to \(d \theta / d t, d a / d t,\) and \(d b / d t\) if none of \(a, b,\) and \(\theta\) are constant?

Short Answer

Expert verified
a) Proportional to \(\cos\theta\cdot\frac{d\theta}{dt}\). b) Depends on \(\frac{da}{dt}\) and \(d\theta/dt\). c) Involves \(\frac{da}{dt}, \frac{db}{dt}, \frac{d\theta}{dt}\).

Step by step solution

01

Understanding the Formula for Area

The given formula for the area of the triangle is \( A = \frac{1}{2} a b \sin \theta \). In this formula, \(a\) and \(b\) are the lengths of the sides, and \(\theta\) is the angle formed between those sides.
02

Differentiating the Area Formula with Respect to Time (a)

Given that both \(a\) and \(b\) are constant, we differentiate the area function with respect to time:\[ \frac{dA}{dt} = \frac{1}{2} a b \cos \theta \cdot \frac{d\theta}{dt} \].This implies that the rate of change of the area is directly proportional to the rate of change of the angle, scaled by a factor \(\frac{1}{2}ab\cos\theta\).
03

Differentiating with Respect to Time when a Changes (b)

Now, \(b\) is constant but \(a\) changes. Differentiate the area:\[ \frac{dA}{dt} = \frac{1}{2} b \sin \theta \cdot \frac{da}{dt} + \frac{1}{2} a b \cos \theta \cdot \frac{d\theta}{dt} \].The rate of change of area depends on both the change in side \(a\) at a rate \(\frac{da}{dt}\) and the change in \(\theta\) at a rate \(\frac{d\theta}{dt}\).
04

Differentiating with Respect to Time when All Variables Change (c)

Differentiate assuming all \(a\), \(b\), and \(\theta\) change:\[ \frac{dA}{dt} = \frac{1}{2} b \sin \theta \cdot \frac{da}{dt} + \frac{1}{2} a \sin \theta \cdot \frac{db}{dt} + \frac{1}{2} ab \cos \theta \cdot \frac{d\theta}{dt} \].Here, the area change rate involves the rates \(\frac{da}{dt}\), \(\frac{db}{dt}\), and \(\frac{d\theta}{dt}\). These terms account for changes in sides \(a\), \(b\), and angle \(\theta\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rate of Change
The rate of change is a fundamental concept in calculus. It describes how a certain quantity changes over time. When dealing with geometric problems, such as the area of a triangle, understanding how changes in angle or side length affect the area can be very insightful.

In our triangle problem, the rates of change correspond to the derivatives of these quantities with respect to time. For instance, if you want to know how quickly the triangle's area is changing, you need to differentiate the area formula with respect to time, which gives you the formula for \(\frac{dA}{dt}\).

  • When sides \(a\) and \(b\) remain constant, \(\frac{dA}{dt} = \frac{1}{2}ab\cos\theta \cdot \frac{d\theta}{dt}\). This shows that the rate of change of the area depends on how fast the angle is changing.
  • If side \(a\) can change, the rate formula includes \(\frac{da}{dt}\), which signifies that the length change also impacts \(\frac{dA}{dt}\).
  • When all sides and the angle can change, \(\frac{dA}{dt}\) encompasses all their rates of change, showing a collective influence on how the area shifts over time.
Triangle Area Formula
The area formula for a triangle, specifically when incorporating trigonometry, is a step beyond the basic \(\frac{1}{2}\times \text{base} \times \text{height}\) formula many of us learn first. For a triangle with sides \(a\) and \(b\), and an enclosed angle \(\theta\), the area is calculated with
\[A = \frac{1}{2}ab\sin\theta \]
This formula provides a way to compute areas when height isn't readily available.
  • The side lengths \(a\) and \(b\) serve as base dimensions of the area.
  • \(\sin\theta\) is critical, as it accounts for how the tilt of side \(a\) relative to \(b\) affects the enclosed space.
  • This form is especially helpful in scenarios involving oblique triangles, which lack a right angle.
Trigonometric Derivatives
Trigonometric derivatives come into play when we need to differentiate trigonometric functions such as sine and cosine. They are foundational in calculus, especially when variables change over time, a common situation in dynamic systems.

For the triangle area problem, the derivative of \(\sin\theta\) is essential. During differentiation of the given area formula:
  • Differentiating \(\sin\theta\) yields \(\cos\theta\cdot\frac{d\theta}{dt}\), indicating the rate at which the angle is affecting the triangle's area.
  • Each side's change and angle change plays a role in the overall differentiation, affecting the formula for \(\frac{dA}{dt}\).
  • These derivatives help translate the geometric changes into manageable mathematical expressions.

Understanding how these trigonometric derivatives link back to the physical dimensions of a problem allows for deeper insight into the world of calculus differentiation, connecting geometry and calculus seamlessly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

A growing sand pile Sand falls from a conveyor belt at the rate of 10 \(\mathrm{m}^{3} / \mathrm{min}\) onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 \(\mathrm{m}\) high? Answer in centimeters per minute.

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. \begin{equation} x^{3}+y^{2}=\sin ^{2} y \end{equation}

A melting ice layer A spherical iron ball 8 in. in diameter is conted with a layer of ice of uniform thickness. If the ice melts at the rate of 10 \(\mathrm{in}^{3} / \mathrm{min}\) , how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0) \end{equation}
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.