/*! 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 3 The angles of a triangle are cal... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 3

The angles of a triangle are calculated from the sides \(a, b, c\). If small changes \(\delta a, \delta b, \delta c\) are made in sides show that approximately \(\delta A=\frac{a}{2 \Delta}\\{\delta a-\delta b \cdot \cos C-\delta c \cdot \cos B\\}\) where \(\Delta\) is the area of the triangle and \(A, B, C\) are the angles opposite to \(a, b, c\) respectively. Verify that \(\delta A+\) \(\delta B+\delta C=0\) Solution: From cosine rule \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\) or \(2 \cos A=\frac{b^{2}+c^{2}-a^{2}}{b c}\) Differentiating both sides, we get \(-2 \sin A \delta A=\) \(\frac{b c\\{2 b \delta b+2 c \delta c-2 a \delta a\\}-\left(b^{2}+c^{2}-a^{2}\right)(b \delta c+c \delta b)}{(b c)^{2}}\) \(-2 b c \sin A . \delta A\) \(=\frac{\left(b^{2} c-c^{3}+a^{2} c\right) \delta b+\left(b c^{2}-b^{3}+a^{2} b\right) \delta c-2 a b c \delta a}{b c}$$\begin{aligned} \Rightarrow &-2 b c\left(\frac{2 \Delta}{b c}\right) . \delta A \\\ &=\frac{c\left(a^{2}-b^{2}-c^{2}\right) \delta b+b\left(a^{2}+c^{2}-b^{2}\right) \delta c-2 a b c \delta a}{b c} \\\ &\\{\Delta=-b c \sin A\\} \\ \Rightarrow &-4 \Delta . \delta A \\\ &=\frac{c(2 a b \cos C) \delta b+b(2 a c \cos B) \delta c-2 a b c \delta a}{b c} \\ \therefore \quad \delta A=\frac{a}{2 \Delta}\\{\delta a-\delta b \cdot \cos C-\delta c \cdot \cos B\\} \\ &\\{\text {... (ii) }\\\ \text { Similarly, } \delta B=\frac{b}{2 \Delta}\\{\delta b-\delta c \cdot \cos A-\delta a \cdot \cos C\\} & \ldots \text { (iii) } \\ \text { and } & \delta C=\frac{c}{2 \Delta}\\{\delta c-\delta a \cdot \cos B-\delta b \cdot \cos A\\} \end{aligned}\) Adding (ii), (iii) and (iv) we get \(\delta A+\delta B+\delta C=\frac{\delta a}{2 \Delta}\\{a-b \cos C-c \cdot \cos B\\}+\) \(\frac{\delta b}{2 \Delta}\\{b-a \cos C-c \cdot \cos A\\}+\) \(\frac{\delta c}{2 \Delta}\\{c-a \cos B-b \cdot \cos A\\}\) \(=\frac{\delta a}{2 \Delta}(a-a)+\frac{\delta b}{2 \Delta}(b-b)+\frac{\delta c}{2 \Delta}(c-c)=0\) Hence \(\delta A+\delta B+\delta C=0\)

Short Answer

Expert verified
The approximate changes in the angles of a triangle with sides a, b, and c and small changes in the sides $\delta a, \delta b, \delta c$ can be represented as: \(\delta A = \frac{a}{2\Delta}(\delta a - \delta b\cos C - \delta c\cos B)\) \(\delta B = \frac{b}{2\Delta}(\delta b - \delta a\cos C - \delta c\cos A)\) \(\delta C = \frac{c}{2\Delta}(\delta c - \delta a\cos B - \delta b\cos A)\) Where $\Delta$ is the area of the triangle and A, B, C are the angles opposite to sides a, b, and c, respectively. The sum of changes in angles $\delta A + \delta B + \delta C = 0$.

Step by step solution

01

Differentiate the cosine rule

The cosine rule states that \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\) To find the change in angle A, differentiate the cosine rule with respect to A: \(-2\sin A \delta A =\frac{bc(2b\delta b + 2c\delta c - 2a\delta a) - (b^2 + c^2 - a^2)(b\delta c + c\delta b)}{(bc)^2}\)
02

Replace \(\sin A\) with expression based on the area

We know that the area of the triangle, denoted by \(\Delta\), is related to the angle and the sides by \(\Delta = \frac{1}{2}bc\sin A\). Hence, \(2\Delta= bc\sin A\). Replace \(\sin A\) in the equation from step 1 by using this expression: \(-2bc\frac{2\Delta}{bc}\delta A =\frac{bc(2b\delta b + 2c\delta c - 2a\delta a) - (b^2 + c^2 - a^2)(b\delta c + c\delta b)}{(bc)^2}\)
03

Simplify and obtain expression for \(\delta A\)

After simplification, we reach \(\delta A = \frac{a}{2\Delta}(\delta a - \delta b\cos C - \delta c\cos B)\)
04

Repeat for angles B and C

Following the same process for angles B and C, we derive similar expressions for the changes in these angles: \(\delta B = \frac{b}{2\Delta}(\delta b - \delta a\cos C - \delta c\cos A)\) \(\delta C = \frac{c}{2\Delta}(\delta c - \delta a\cos B - \delta b\cos A)\)
05

Verify that \(\delta A + \delta B + \delta C = 0\)

To verify that the sum of changes in angles equals zero, we add the expressions for \(\delta A\), \(\delta B\), and \(\delta C\): \(\delta A + \delta B + \delta C = \frac{\delta a}{2\Delta}(a - b\cos C - c\cos B) + \frac{\delta b}{2\Delta}(b - a\cos C - c\cos A) + \frac{\delta c}{2\Delta}(c - a\cos B - b\cos A)\) After simplification, we get: \(\delta A + \delta B + \delta C = \frac{\delta a}{2\Delta}(a-a) + \frac{\delta b}{2\Delta}(b-b) + \frac{\delta c}{2\Delta}(c-c) = 0\) Thus, we have verified that \(\delta A + \delta B + \delta C = 0\).

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Key Concepts

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

Cosine Rule Differentiation

Differential calculus allows us to find the rate of change of quantities, which is vital in understanding the behavior of geometric figures as their elements vary. Specifically, when analyzing how the angles of a triangle change with respect to its sides鈥攁n indispensable concept in triangle geometry鈥攊s where the cosine rule comes into play. The cosine rule links the lengths of the sides of a triangle to the cosine of one of its angles.


Upon differentiating the cosine rule with respect to one side, we can derive the rate at which the corresponding angle changes. This rate, noted as \(\delta A\), \(\delta B\), or \(\delta C\), for angles A, B, and C, respectively, provides insight into how infinitesimal changes in the triangle's sides impact its angles.

  • The differentiation of the cosine rule helps us study the sensitivity of triangle angles to changes in sides.
  • It involves utilizing derivatives to understand the dynamic relationship between angles and sides in non-static scenarios.
Relationship Between Triangle Sides and Angles

The geometry of triangles is governed by several foundational relationships among its angles and sides. The cosine rule is one such relationship that connects the lengths of a triangle's sides to the cosine of the opposite angles. The rule asserts that for a triangle with sides \(a\), \(b\), and \(c\), and angle A opposite to side \(a\), the relationship \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\) holds true.


This relationship becomes particularly relevant when considering small changes in the sides. Knowing how angle A adjusts as side \(a\) undergoes infinitesimal changes鈥攃aptured by differentiation of the cosine rule鈥攊s critical for applications in physics, engineering, and even advanced mathematics. Understanding these concepts can eliminate confusion in situations where precision matters, and slight variations could lead to significant results.

Area of a Triangle Calculation

Calculating the area of a triangle is a fundamental task in geometry. Commonly, we use the formula \(\Delta = \frac{1}{2} bc \sin A\), where \(\Delta\) represents the area, and \(A\), \(b\), and \(c\) are one angle and the lengths of the sides forming that angle, respectively. This formula showcases the direct dependence of the triangle's area on its sides and angles, particularly the sine of the included angle.


When dealing with calculus problems that involve variations in side lengths, such as \(\delta a\), \(\delta b\), and \(\delta c\), it is essential to consider the corresponding effect on the area, since the two quantities are intertwined. For students tackling geometric optimization problems or studying the properties of triangles, mastering the calculation of a triangle's area using the sine function is vital.

Infinitesimal Changes in Geometric Figures

In differential calculus, infinitesimal changes refer to exceedingly small variations in a quantity. When we apply this concept to geometric figures, such as triangles, we examine the minutest changes in sides or angles and their consequential effects on the figure's properties. Particularly, the understanding of how these infinitesimal changes in sides impact angles鈥攁s delineated by the differentiation of the cosine rule鈥攔eveals a nuanced view of the triangle's stability and how it might deform.


When evaluating these infinitesimal changes, it's important to recall that changes in a triangle's angles resulting from the alteration of sides are interconnected, as evidenced by the fact that the sum of the changes in the angles \(\delta A + \delta B + \delta C\) equals zero. This result underlines a fundamental property of triangles: even with slight fluctuations in their sides, the internal angles adjust to preserve the triangle's integrity.

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Most popular questions from this chapter

Determine a differentiable function \(y=f(x)\) which satisfies \(f^{\prime}(x)=[f(x)]^{2}\) and \(f(0)=-\frac{1}{2} .\) Find also the equation of the tangent at the point where to curve crosses the \(y\)-axis. Solution: \(f(x)=[f(x)]^{2}\) $$ \begin{aligned} &\frac{f^{\prime}(x)}{[f(x)]^{2}}=1 \\ &\operatorname{put} f(x)=t \Rightarrow f^{\prime}(x)=d t \\ &-\frac{1}{f(x)}=x+c \\ &f(x)=-\frac{1}{x+c} \\ &x=0 \Rightarrow f(0)=-\frac{1}{c} \\ &-\frac{1}{2}=-\frac{1}{c} \Rightarrow c=2 \\ &\text { then } f(x)=-\frac{1}{x+2} \\ &y--\left.\frac{1}{x+2} \Rightarrow \frac{d y}{d x}\right|_{\left(0, \frac{1}{2}\right)}=\frac{1}{4} \end{aligned} $$ curve crosses \(y\)-axis i.e., at \(y\)-axis \(x=0\) then \((x, y)\) $$ =\left(0,-\frac{1}{2}\right) $$ Equation of tangent \(\left(y+\frac{1}{2}\right)=\frac{1}{4}(x-0)\) $$ \begin{aligned} \frac{2 y+1}{2} &=\frac{x}{4} \\ x-4 y &=2 \text { Ans } \end{aligned} $$

A curve \(C\) has the property that if the tangent drawn at any point \(P\) on \(C\) meets the co-ordinate axes at \(A\) and \(B\), then \(P\) is the mid-point of \(A B\). The curve passes through the point \((1,1)\). The equation of the curve is (a) \(x y=1\) (b) \(y^{2}=2 x-1\) (c) \(x^{2}=2 y-1\) (d) None of these

The tangent to curve \(y=x-x^{3}\) at point \(P\) mects the curve again at \(Q\). Prove that one point of trisection of \(P Q\) lies on \(y\)-axis. Find locus of other point of trisection Solution: \(\quad\left(y-k_{1}\right)=\left(1-3 h_{1}^{2}\right)\left(x-h_{1}\right)\) \(\left(k_{2}-k_{1}\right)=\left(1-3 h_{1}^{2}\right)\left(h_{2}-h_{1}\right)\) \(\left(h_{2}-h_{2}^{3}-h_{1}+h_{1}^{3}\right)=\left(1-3 h_{1}^{2}\right)\left(h_{2}-h_{1}\right)\) \(\left(h_{2}-h_{1}\right)-\left(h_{2}-h_{1}\right)\left(h_{2}^{2}+h_{1}^{2}+h_{1} h_{2}\right)=\left(1-3 h_{1}^{2}\right)\left(h_{2}-h_{1}\right)\) \(1-h_{2}^{2}-h_{2}^{2}-h_{1} h_{2}=1-3 h_{1}^{2}\) \(2 h_{1}^{2}-h_{2}^{2}-h_{1} h_{2}=0\) \(2 h_{1}^{2}-2 h_{1} h_{2}+h_{1} h_{2}-h_{2}^{2}=0$$\left(h_{1}-h_{2}\right)\left(2 h_{1}+h_{2}\right)=0\) \(h_{1}=h_{2 / 2}\) \(M_{1}\left(\frac{h_{2}+2 h_{1}}{3}, \frac{k_{1}+2 k_{1}}{3}\right)\) \(M_{2}\left(\frac{h_{1}+2 h_{2}}{3}, \frac{k_{1}+2 k_{2}}{3}\right)\) \(k=\frac{h_{2}}{h}=-h_{1}\) \(k=\frac{3 h_{1}+15 h_{1}^{3}}{3}=h-5 h^{3}\) or \(y=x-5 x^{3}\)

Find the condition that the line \(x \cos \alpha+y \sin \alpha-p\) may touch the curve \(\left(\frac{x}{a}\right)^{m}+\left(\frac{y}{b}\right)^{m}=1\) Solution: Given equation: \(\left(\frac{x}{a}\right)^{\prime \prime \prime}+\left(\frac{y}{b}\right)^{m}=1\) Differentiating the equation of curve with respect to \(x\) we get, \(\mathrm{m}\left(\frac{x}{a}\right)^{m-1}+m\left(\frac{y}{b}\right)^{m-1} \frac{1}{b} \frac{d y}{d x}=0\) on simplifying we get \(\frac{d y}{d x}=\frac{-b^{m} x^{m-1}}{a^{m} y^{m-1}}\) Hence, at any point \(p\left(x_{1}, y_{1}\right)\) on the curve, Slope of tangent \(=\left(\frac{d y}{d x}\right)_{(x, y)}=\frac{-b^{m \prime} x_{1}^{m-1}}{a^{m} y_{1}^{m-1}}\) \(\therefore\) Equation of tangent at \(p\) is \(y-y_{1}=\frac{-b^{m} x_{1}^{m-1}}{a^{m} y_{1}^{m-1}}\left(x-x_{1}\right)$$\Rightarrow \frac{y y_{1}^{m-1}}{b^{m}}-\frac{y_{1}^{m}}{b^{m}}=-\frac{x x_{1}^{m-1}}{a^{m}}+\frac{x_{1}^{m}}{a^{m}}\) i.e., \(\frac{x}{a}\left(\frac{x_{1}}{a}\right)^{m-1}+\frac{y}{b}\left(\frac{y_{1}}{b}\right)^{m-1}\) \(=\left(\frac{x_{1}}{a}\right)^{m}+\left(\frac{y_{1}}{b}\right)^{m}=1\) (since \(p\) lies on the curve at any point) Hence, the equation of tangent at \(P\left(x_{1}, y_{1}\right)\) on the curve is, \(\frac{x}{a}\left(\frac{x_{1}}{a}\right)^{m-1}+\frac{y}{b}\left(\frac{y_{1}}{b}\right)^{m-1}=1\) and \(x \cos \alpha+y \sin \alpha=P\) If equation (ii) is the equation of tangent, then coefficients of (i) and (ii) must be proportional for point \(\left(x_{1}, y_{1}\right)\) i.c., \(\frac{\cos \alpha}{\frac{1}{a}\left(\frac{x_{1}}{a}\right)^{m-1}}=\frac{\sin \alpha}{\frac{1}{b}\left(\frac{y_{1}}{b}\right)^{m-1}}=\frac{P}{1}\) This gives \(\frac{x_{1}}{a}=\left(\frac{a \cos \alpha}{P}\right)^{\frac{1}{m-1}}\); \(\frac{y_{1}}{b}=\left(\frac{b \sin \alpha}{P}\right)^{\frac{1}{m-1}}\)Since point \(P\left(x_{1}, y_{1}\right)\) lies on the curve, \(\left(\frac{x_{1}}{a}\right)^{m}+\left(\frac{y_{1}}{b}\right)^{m}=1\) i.c. \(\left(\frac{a \cos \alpha}{p}\right)^{\frac{m}{m-1}}+\left(\frac{b \sin \alpha}{p}\right)^{\frac{m}{m-1}}=1\) i.e., \((a \cos \alpha)^{\frac{m}{m-1}}+(b \sin \alpha)^{\frac{1}{m-1}}=\frac{m}{p^{m-1}} .\) This is the required condition.

Show that the distance from the origin of the normal at any point of the curve \(x=a \mathrm{e}^{8}\left(\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2}\right)\) and \(y=a e^{\theta}\left(\cos \frac{\theta}{2}-2 \sin \frac{\theta}{2}\right)\) is twice the distance of the tangent at the point from the origin. Solution: \(\quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\) \(=\frac{a\left[e^{\theta}\left(-\sin \frac{\theta}{2} \cdot \frac{1}{2}-2 \cos \frac{\theta}{2} \cdot \frac{1}{2}\right)+\left(\cos \frac{\theta}{2}-2 \sin \frac{\theta}{2}\right) e^{\theta}\right]}{a\left[e^{\theta}\left(\cos \frac{\theta}{2} \cdot \frac{1}{2}-2 \sin \frac{\theta}{2} \cdot \frac{1}{2}\right)+\left(\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2}\right) e^{\theta}\right]}\) \(=\frac{e^{\theta}\left[-\frac{1}{2} \sin \frac{\theta}{2}-\cos \frac{\theta}{2}+\cos \frac{\theta}{2}-2 \sin \frac{\theta}{2}\right]}{e^{\theta}\left[\frac{1}{2} \cos \frac{\theta}{2}-\sin \frac{\theta}{2}+\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2}\right]}\) \(=-\frac{5 / 2 \sin \theta / 2}{5 / 2 \cos \theta / 2}=-\tan \frac{\theta}{2}\) Equation of normal \(\Rightarrow y-a e^{\theta}\left(\cos \frac{\theta}{2}-2 \sin \frac{\theta}{2}\right)\) \(=\frac{\cos \theta / 2}{\sin \theta / 2}\left(x-a e^{o}\left(\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2}\right)\right)\) \(\Rightarrow y \cdot \sin \frac{\theta}{2}-a e^{\theta} \cos \frac{\theta}{2} \sin \frac{\theta}{2}+2 a e^{\theta} \sin ^{2} \frac{\theta}{2}=x \cos \frac{\theta}{2}\) \(-a \mathrm{e}^{0} \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}-2 a e^{\theta} \cos ^{2} \frac{\theta}{2}\) \(\Rightarrow x \cos \frac{\theta}{2}-y \sin \frac{\theta}{2}=2 a e^{\theta}\) \(\Rightarrow x \cos \frac{\theta}{2}-y \sin \frac{\theta}{2}-2 a e^{\theta}=0\) its distance from origin \(p_{1}=\left|\frac{0+0-2 a e^{\theta}}{\sqrt{\cos ^{2} \frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}}}\right|\) \(p_{1}=2 a e^{\theta}\) Similarly find equation of tangent and solve.
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