What is the slope of the line normal to the tangent line of #f(x) = xcotx+2xsin(x-pi/3) # at # x= (15pi)/8 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = xcotx+2xsin(x-pi/3) # at # x= (15pi)/8  #?

Cameron Alexander · Accepted Answer

#-2/(15sqrt(3)pi + 4)#

#rArr xcotx + 2xsin(x - pi/3)#

For reference :- #sin ((15pi)/8) = 1/2# #cos ((15pi)/8) = -sqrt3/2# #cosec ((15pi)/8) = 2# #cot ((15pi)/8) = -sqrt3#

As slope of any line is the derivative of the equation, I will differentiate the equation of tangent.

#rArr dy/dx = -xcosec^2xcotx + 2[sin(x - pi/3) + cos(x - pi/3)]#

clearly, Here I can see the trigonometric formula of #sin(A - B) and cos(A - B)# Solving, #sin(x - pi/3) + cos(x - pi/3)# by above 2 formulas, I get, #(sinx + cosx) + sqrt3(sinx - cosx)#

So, at #x = (15pi)/8#, The value of above equation becomes #2#

Now, at #x = (15pi)/8#, #dy/dx = (15sqrt(3)pi + 4)/2 = m_"1"#

Now, when a line is normal to the tangent line, it means that the line is perpendicular to the tangent line.

Let the slope of perpendicular line be #m_"2"#

If the lines are perpendicular to each other then, #m_"1"m_"2" = -1#

So, using above equation the slope of line normal to the tangent comes out to be,

#m_"2" = -2/(15sqrt(3)pi + 4)#

ENJOY MATHS !!!!!!!!

Aurora Alvarado · Answer

undefined To find the slope of the line normal to the tangent line of $f(x) = x\cot(x) + 2x\sin(x-\frac{\pi}{3})$ at $x = \frac{15\pi}{8}$, we first find the derivative of $f(x)$ using the product and chain rule. Then, we evaluate the derivative at $x = \frac{15\pi}{8}$ to get the slope of the tangent line. Finally, we take the negative reciprocal of this slope to find the slope of the line normal to the tangent line.