How do you find an equation of the tangent line to the curve at the given point #y = sec (x) - 6 cos (x)# and #P= (pi/3, -1)#?

Question

Ezekiel Alexander · Accepted Answer

# y = 5sqrt(3)x - 5sqrt(3)*pi/3 - 1#

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is #−1#).

We have:

# y = secx-6cosx #

Differentiating wrt #x# we get

# dy/dx = secxtanx+6sinx #

When #x=pi/3#; we have:

# y = sec(pi/3)-6cos(pi/3) #
# \ \ = 2-3 #
# \ \ = -1 => P# lies on curve
And:

# dy/dx = sec(pi/3)tan(pi/3)+6sin(pi/3) #
# \ \ \ \ \ \ = 2sqrt(3)+6sqrt(3)/2 #
# \ \ \ \ \ \ = 5sqrt(3) #

So using the point/slope form #y-y_1=m(x-x_1)# the tangent equations we seek is;

# y - (-1) = 5sqrt(3) ( x - pi/3 ) #
# :. y +1 = 5sqrt(3)x - 5sqrt(3)*pi/3#
# :. y = 5sqrt(3)x - 5sqrt(3)*pi/3 - 1#

We can confirm this graphically:

William Abdalla · Answer

undefined To find the equation of the tangent line to the curve at the given point P = (pi/3, -1), we need to find the derivative of the function y = sec(x) - 6cos(x) and evaluate it at x = pi/3.

The derivative of y = sec(x) - 6cos(x) can be found using the chain rule and product rule. The derivative is given by:

dy/dx = sec(x)tan(x) + 6sin(x)

Evaluating this derivative at x = pi/3, we get:

dy/dx = sec(pi/3)tan(pi/3) + 6sin(pi/3)

Simplifying further, we have:

dy/dx = (2/√3)(√3/3) + 6(√3/2)

dy/dx = 2/3 + 3√3

Now, we have the slope of the tangent line at the point P. Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values of x1 = pi/3, y1 = -1, and m = 2/3 + 3√3, we get:

y - (-1) = (2/3 + 3√3)(x - pi/3)

Simplifying further, we have:

y + 1 = (2/3 + 3√3)(x - pi/3)

This is the equation of the tangent line to the curve y = sec(x) - 6cos(x) at the point P = (pi/3, -1).