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)#?

1 Answer
Steve M
Aug 4, 2017

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

Explanation:

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:
enter image source here

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