Question

To find the acute angle between the curves and point of intersection.

Short answer

The point of intersection is \(\left( {\frac{\pi }{4},\frac{{\sqrt 2 }}{2}} \right)\). The angle between the curves is \({70.5^\circ }\).

Step-by-step solution

Concept of Formula Used

Formula:

Write the expression for line equation.

\(y = mx + c\)

Here,

\(c\)is constant, and

\(m\)is slope.

Write the expression to find\({\rm{a}} \cdot {\rm{b}}\)in terms of\(\theta \).

\({\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta \)

Here,

\(|{\rm{a}}|\)is the magnitude of a vector,

\(|{\rm{b}}|\)is the magnitude of b vector, and

\(\theta \)is the angle between vectors\(a\)and\(b\).

Rearrange equation.

\(\theta = {\cos ^{ - 1}}\left( {\frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}} \right)\)

Calculation for the point of intersection

The given equations are,

\(y = \sin x(1)\)

\(y = \cos x(2)\)

Substitute equation \(\left( 1 \right)\)in equation \(\left( 2 \right).\)

\(\sin x = \cos x\)

Rearrange the equation.

\(\begin{aligned}{l}\frac{{\sin x}}{{\cos x}} &= 1\\\tan x &= 1\\x &= {\tan ^{ - 1}}(1)\\x &= \frac{\pi }{4}\end{aligned}\)

At,\(0 \le x \le \frac{\pi }{2}\)

In equation\(\left( 1 \right)\), substitute \(\frac{\pi }{4}\) for \(x\).

\(\begin{aligned}{l}dy &= \sin \left( {\frac{\pi }{4}} \right)\\dy &= \frac{{\sqrt 2 }}{2}\end{aligned}\)

In equation\(\left( 2 \right)\)\(\frac{\pi }{4}\)for \(x\).

\(\begin{aligned}{l}dy &= \cos \left( {\frac{\pi }{4}} \right)\\dy &= \frac{{\sqrt 2 }}{2}\end{aligned}\)

Thus, the point of intersection is \(\left( {\frac{\pi }{4},\frac{{\sqrt 2 }}{2}} \right)\)

Solve for the vectors\(a\)and\(b\)

Differentiate equation \(\left( 1 \right)\)with respect to \(x\).

\(\begin{aligned}{l}\frac{d}{{dx}}(y) &= \frac{d}{{dx}}(\sin x)\\\frac{d}{{dx}}(y) &= \cos x\end{aligned}\)

Substitute \(\frac{\pi }{4}\) for \(x\).

\(\begin{aligned}{l}\frac{d}{{dx}}(y) &= \cos \left( {\frac{\pi }{4}} \right)\\\frac{d}{{dx}}(y) &= \frac{{\sqrt 2 }}{2}\end{aligned}\)

Differentiate equation \(\left( 2 \right)\)with respect to \(x\).

\(\begin{aligned}{l}\frac{d}{{dx}}(y) &= \frac{d}{{dx}}(\cos x)\\\frac{d}{{dx}}(y) &= - \sin x\end{aligned}\)

Substitute \(\frac{\pi }{4}\) for \(x\).

\(\begin{aligned}{l}\frac{d}{{dx}}(y) &= - \sin \left( {\frac{\pi }{4}} \right)\\\frac{d}{{dx}}(y) &= - \frac{{\sqrt 2 }}{2}\end{aligned}\)

So, the tangent lines at the point \((1,1)\) have slopes \(\frac{{\sqrt 2 }}{2}\) and \( - \frac{{\sqrt 2 }}{2}\).

The vectors parallel to the tangent lines are \(\left\langle {1,\frac{{\sqrt 2 }}{2}} \right\rangle \) and \(\left\langle {1, - \frac{{\sqrt 2 }}{2}} \right\rangle \).

Calculation for the dot product\({\rm{a}} \cdot {\rm{b}}\)

Consider \({\rm{a}} = \left\langle {1,\frac{{\sqrt 2 }}{2}} \right\rangle \) and \({\rm{b}} = \left\langle {1, - \frac{{\sqrt 2 }}{2}} \right\rangle \).

Consider a general expression to find dot product between two two-dimensional vectors.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2}} \right\rangle \cdot \left\langle {{b_1},{b_2}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2}\end{aligned}\)

In above equation, substitute \(1\) for \({a_1},\frac{{\sqrt 2 }}{2}\) for \({a_2},1\) for \({b_1}\) and \( - \frac{{\sqrt 2 }}{2}\) for \({b_2}\).

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (1)(1) + \left( {\frac{{\sqrt 2 }}{2}} \right)\left( { - \frac{{\sqrt 2 }}{2}} \right)\\{\rm{a}} \cdot {\rm{b}} &= 1 - \frac{2}{4}\\{\rm{a}} \cdot {\rm{b}} &= 1 - \frac{1}{2}\\{\rm{a}} \cdot {\rm{b}} &= \frac{1}{2}\end{aligned}\)

Calculation for the magnitude of\(a\)and\(b\)

Consider a general expression to find magnitude of a two dimensional vector that is\(a = \left\langle {{a_1},{a_2}} \right\rangle \).\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2} \)

Similarly, Consider a general expression to find magnitude of a two dimensional vector that is \({\rm{b}} = \left\langle {{b_1},{b_2}} \right\rangle \).

\(|{\rm{b}}| = \sqrt {b_1^2 + b_2^2} \)

In above equation, substitute \(1\) for \({a_1}\) and \(\frac{{\sqrt 2 }}{2}\) for \({a_2}\).

\(\begin{aligned}{l}|{\rm{a}}| &= \sqrt {{{(1)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}} \\|{\rm{a}}| &= \sqrt {1 + \frac{2}{4}} \\|{\rm{a}}| &= \sqrt {1 + \frac{1}{2}} \\|{\rm{a}}| &= \sqrt {\frac{3}{2}} \end{aligned}\)

In above equation, substitute \(1\) for \({b_1}\) and \( - \frac{{\sqrt 2 }}{2}\) for \({b_2}\).

\(\begin{aligned}{l}|{\rm{b}}| &= \sqrt {{{(1)}^2} + {{\left( { - \frac{{\sqrt 2 }}{2}} \right)}^2}} \\|{\rm{b}}| &= \sqrt {1 + \frac{2}{4}} \\|{\rm{b}}| &= \sqrt {1 + \frac{1}{2}} \\|{\rm{b}}| &= \sqrt {\frac{3}{2}} \end{aligned}\)

Calculation for the acute angle between the curves

In formula, substitute \(\frac{1}{2}\) for \(a.b,\sqrt {\frac{3}{2}} \) for \(|a|\)and \(\sqrt {\frac{3}{2}} \) for \(\left| b \right|.\)

\(\begin{aligned}{l}\theta &= {\cos ^{ - 1}}\left( {\frac{{\left( {\frac{1}{2}} \right)}}{{\sqrt {\frac{3}{2}} \sqrt {\frac{3}{2}} }}} \right)\\\theta &= {\cos ^{ - 1}}\left( {\frac{{\left( {\frac{1}{2}} \right)}}{{\left( {\frac{3}{2}} \right)}}} \right)\\\theta &= {\cos ^{ - 1}}\left( {\frac{1}{3}} \right)\\\theta &= {70.5^\circ }\end{aligned}\)

Thus, the acute angle between the curves is \({70.5^\circ }\).