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CHAPTER 10
10.23
Find an equation of the normal line to the curve y = 1 + cos x at the point
The slope of the tangent line is the derivative y'. But, y' = -sin x = -sin (ir/3) = -V3/2. Hence, the
slope of the normal line is the negative reciprocal 2A/3. So a point-slope equation of the normal line is
y - \ = (2/V3)(x - IT/3) = (2V3/3)jc - 27rV3/9.
10.24
Derive the formula
Remember that tan x = sin AT/COS x and sec x = 1 /cos x. By the quotient rule,
10.25 Find an equation of the tangent line to the curve y = tan
2 x at the point (7r/3,3).
Note that tan (ir/3) = sin(7j-/3)/cos (w/3) = (V3/2)/i = V3, and sec(ir/3) = l/cos(ir/3) = 1/| =2. By
the chain rule, / = 2(tanx)- -7- (tan*) = 2(tan*)(sec
2 *). Thus, when x = ir/3, y' =2V5-4 = 8V5, so
the slope of the tangent line is 8V3. Hence, a point-slope equation of the tangent line is y — 3 = 8V3(x - ir/3).
10.26
Derive the formula
By the identity cot x = tan (IT12 - x) and the chain rule,
10.27
Show that
By the chain rule,
10.28
Find an equation of the normal line to the curve y = 3 sec
2 x at the point (ir/6,4)
10.29
Find D x
By the chain rule, y' = 3[2 sec x • -r- (sec x)] = 3(2 sec x • sec x • tan x) = 6 sec
2 x tan x. So the slope of the
tangent line is y' = 6(f)(V3/3) = 8V3/3. Hence, the slope of the normal line is the negative reciprocal
-V3/8. Thus, a point-slope equation of the normal line is y - 4 = -(V3/8)(x - 77/6).
Recall that D,(csc;c) = -esc x cot x. Hence, by the chain rule,
10.30
Evaluate
10.31 Evaluate
Hence,