Question

A) As \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R\) \(\Rightarrow a=2 R \sin A\) \(\Rightarrow \frac{d a}{d A}=2 R \cos A\) \(\Rightarrow \frac{d a}{\cos A}=2 R d A\) Similarly, \(\frac{d b}{\cos B}=2 R d B, \frac{d c}{\cos c}=2 R d C\) \(\quad \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}+1=\) \(2 R(d A+d B+d C)+1\) \(\quad A s A+B+C=\pi\) \(\Rightarrow d A+d B+d C=0\) using eq (I) \(|\mathbf{m}|=1\) B) $\begin{aligned} & \Rightarrow \mathrm{m}=\pm 1 \\ \mathrm{x}^{2} \mathrm{y}^{2}=& 16 \end{aligned}$ \(2 x^{2} y \frac{d y}{d x}+2 x y^{2}=0\) \(\frac{d y}{d x}=-\frac{y}{x}=1\) Subtangent \(=y / d y / d x=2=|\mathrm{k}|\) \(\Rightarrow \mathrm{k}=\pm 2\) C) \(y=2 e^{2 x}\) \(\frac{d y}{d x}=4 e^{2 x}=4\) for yaxis $\tan ^{-1} 4=\frac{\pi}{2}-\cot ^{-1}\left(\frac{8 n-4}{3}\right)=\tan ^{-1}\left(\frac{8 n+4}{3}\right)$ \(12=|8 n-4|\) \(\mathrm{n}=2\) or \(-1\) D) \(\mathrm{x}=\underline{\mathrm{e}^{\sin y}}\) \(1=e^{\sin y} \cos y \frac{d y}{d x}\) At \((1,0) \Rightarrow \frac{d y}{d x}=1\) eqn of Normal \(\rightarrow y=-(x-1)\) \(y+x-1=0\) Area of \(\Delta=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\) \(\Rightarrow|2 t+1|=3\) \(2 t+1=\pm 3 \Rightarrow t=1\) or \(-2\) $\mathrm{A} \rightarrow(\mathrm{PQ}), \mathrm{B} \rightarrow(\mathrm{RS}), \mathrm{C} \rightarrow(\mathrm{RQ}), \mathrm{D} \rightarrow(\mathrm{PS})$

Short answer

Answer: The relationship between the variables and their rates of change is given by \(|\boldsymbol{m}|=1\). The subtangent of the curve is 2. The angle between the tangent of the curve and the y-axis is given by \(\tan^{-1}(4) = \tan^{-1}\left(\frac{8n+4}{3}\right)\). The equation of the normal of the curve is \(y = -(x-1)\), and the area of the triangle is \(\frac{1}{2}\).

Step-by-step solution

Derive equations

From the given equation, we have \(a = 2R\sin A\), \(b = 2R\sin B\), and \(c = 2R\sin C\). Differentiate with respect to the angles to obtain: \(\frac{da}{dA} = 2R\cos A\), \(\frac{db}{dB} = 2R\cos B\), and \(\frac{dc}{dC} = 2R\cos C\).

Combine equations and obtain the result

Now we can rewrite the equations as \(\frac{d a}{\cos A}=2 R d A\), \(\frac{d b}{\cos B}=2 R d B\), and \(\frac{d c}{\cos C}=2 R d C\). Adding these equations and use the fact that \(dA + dB + dC = 0\), we get \(|\boldsymbol{m}|=1\). #Part B: Find the subtangent of a curve#

Write down the given equation

We have the equation \(xy^2 = 16\).

Differentiate the equation

Differentiate the equation with respect to \(x\) to obtain \(\frac{dy}{dx} = -\frac{y}{2x}\).

Find the subtangent

The subtangent is given by \(|\boldsymbol{k}| = \frac{y}{\frac{dy}{dx}} = 2\). #Part C: Angle between the tangent and y-axis#

Write down the given equation

We have the equation \(y = 2e^{2x}\).

Differentiate the equation

Differentiate the equation with respect to \(x\) to obtain \(\frac{dy}{dx} = 4e^{2x}\).

Find the angle

At the y-axis, substitute \(x=0\) to find \(\frac{dy}{dx} = 4\). The angle between the tangent and the y-axis is given by: \(\tan^{-1}(4) = \tan^{-1}\left(\frac{8n+4}{3}\right)\). #Part D: Equation of the normal and the area of the triangle#

Write down the given equation

We have the equation \(x = e^{\sin y}\).

Differentiate the equation

Differentiate the equation with respect to \(x\) to obtain \(\frac{dy}{dx} = \frac{1}{e^{\sin y}\cos y}\).

Find the equation of normal line

At \((1, 0)\), substitute the values to find \(\frac{dy}{dx} = 1\). The equation of normal line is given by: \(y = -(x-1)\).

Find the area of the triangle

Area of the triangle is given by \(\Delta = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\). To find the values of \(t\), we can simply solve the equation \(|2t+1| = 3\) to obtain \(t=1\) or \(-2\).