Calculus problem #2153

Before solving this problem, one must recognize two key notes:

1. >(From Details and assumptions )Two curves intersect at a right angle if the tangent lines of the curves at the intersection point are perpendicular to each other.

2. From basic algebra, it is known that if two slopes, $m_{1}$ and $m_{2}$ , are perpendicular to each other, the statement, $m_{1}$ = $\frac{-1}{m_2}$ , must be true .

To implement Calculus, we must recognize that the derivative of a function at a point $x$ is the slope of the tangent line at that point.

Step 1: Take the derivatives of both equations.

Equation 1:

$y_1 = \ln (2x + 3)$

Applying the Log Rule from Differentiation of Algebraic Functions and the Chain Rule from Rules of Differentiation

$y_1^{'} = \frac{1}{2x+3}\frac{d(2x+3)}{dx}$ $\Rightarrow$ $y_1^{1} = \frac{1}{2x+3}(2)$ $\Rightarrow$ $y_1^{'} = \frac{2}{2x+3}$

Equation 2:

$y_2 =a - \ln x$

Only differentiating $-\ln x$ because the derivative of $a$ , a constant, is $0$ .

$y_2^{'} = \frac{-1}{x}$

Step 2: Find an equation to relate the perpendicular nature of the two functions at point of intersection, $x$

Letting $m_1 = y_1^{'} = \frac{2}{2x+3}$ and $m_2 = y_2^{'} = \frac{-1}{x}$

Referring to key note #2 for perpendicular slopes and plugging in:

$m_{1}$ = $\frac{-1}{m_2}$

$\Rightarrow \frac{2}{2x+3}$ = $\frac{-1}{(\frac{-1}{x})}$

$\Rightarrow \frac{2}{2x+3} = x$

Multiplying both sides by $(2x+3)$ :

$\Rightarrow 2 = (2x+3)x$

Distributing the $x$ on the RHS:

$\Rightarrow 2 = 2x^2 +3x$

Subtracting 2 from both sides:

$\Rightarrow 2x^2 +3x -2 = 0$

Factoring the quadratic equation:

$\Rightarrow (x+2)(2x-1) = 0$

Setting each factor $=0$ and solving for $x$ :

$x+2 = 0 \hspace{1 in} 2x-1 = 0$

$\Rightarrow \boxed{x = -2} \hspace{1 in} \Rightarrow \boxed{x = \frac{1}{2}}$

Step 3: Understanding that because the two equations intersect, they share a common $x$ and $y$ value, which can be found by testing both $x$ 's for extraneous solutions:

Testing with equation 1 because it contains no unknown constants, such as, $a$ , in equation 2:

$x = -2: \hspace{1 in} x = \frac{1}{2}:$

$\Rightarrow y = \ln (2(-2) + 3) \hspace{1 in} \Rightarrow y = \ln(2(\frac{1}{2}) + 3)$

$\Rightarrow y = \ln -1 \hspace{10 mm}$ or $\hspace{ 10 mm}\Rightarrow y = \ln 4$

Realizing that $y$ can't be $\ln -1$ , because $y\in\mathbb{R}$ and $\ln -1$ is an imaginary number.

Thus, $\boxed{x = \frac{1}{2}}$ and therefore $\boxed {y = \ln 4}$

Step 4: Use found $(x,y)$ to find $a$ in Equation 2:

$\ln 4 = a - \ln \frac{1}{2}$

Adding $\ln \frac{1}{2}$ to both sides:

$\Rightarrow \ln 4 + \ln \frac{1}{2} = a$

Multiplying the inputs of the logarithms as a use of the Product Rule of Logarithms

$\Rightarrow \ln(4(\frac{1}{2})) = a$

Therefore, $\boxed{a = \ln 2 \approx 0.693}$

Whenever two curves are perpendicular to each other then their slopes are connected to each other by the following relation:

(slope of 1st curve) times (slope of 2nd curve) = -1

Now finding out their slopes by differentiating their respective curve functions, and putting into the above relation, we find 2 values of x, x = 1/2 and -2, out of which -2 is invalid(by putting the x = -2 in y = ln(2x+3), we get ln(-1) which is obviously not good in mathematics, you know. :) ) Now putting x = 1/2 in the equation y = ln(2x+3) , we get y = 1.386. Now putting the values of x and y in the 2nd equation y = a - ln x , we get the value of a which is .693.

Thank you for considering my solution.

y = ln(2x+3), dy/dx = 2/(2x+3)

y = a - lnx, dy/dx = -1/x

as, tangent lines of the curves at the intersection point are perpendicular to each other.

{2/(2x+3) }x{-1/x} = -1

2x^2 + 3x -2 =0

x = 1/2 & -2

put x = 1/2 on following equation,

ln(2x+3) = a - lnx

a = ln4 + ln2

a= 0.69314718