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At x = 3 the equation of tangent to y = x3 is : z3(x) = 27x − 54
At x = 4 the equation of tangent to y = x 3 is : z4(x) = 48x − 128
Excercise 1 : Draw the graph of y = x 3 and complete the following table. tanθi y’(xi) y(xi) 3.078 3 3 3x − 2 o 12 8 12x − 16 87 o 27 27 27x − 54 88 o 48 64 48x − 128 xi approx. θi 1 72 o 2 85 3
4 101 zi (x) = y ’(xi ) . x + c i /
Exercise 2: Repeat exercise 1 with y(x) = 1 2 x2. What if tan θ = 0 ? , that is to say the tangent to the curve is parallel to the
x-axis. We shall deal with this in the next few chapters. Exercise 3: In our main example of the projectile depicted below with u = 2√2
and θ = π/4 draw the graph and complete the table as in exercise 1 for
instants t1 = T/ 4 , t2 = T/ 2 , and t3 = 3T/ 4 .
height y(t) = u . s i n θ . t -- 1 g t 2
z 3 (t 3 ) = y(t 3 ) ( θ3 (0,0) T/ 2 t3 T t y’(t 3 ) = tan( θ3 )
y ’(t) = u . s i n θ -- g t tangent
It does not make sense to talk about a tangent to a WELL-BEHAVED linear function.
However, its SLOPE = FIRST DERIVATE = tan θ, when θ is the angle of inter section
the linear function makes with the x-axis.
With the concept of the tangent in mind, the reader may wish to review the bouncing
ball e xample on page 91 and gain an Anal ytical g eometr y v iew of
diff erentia bility.
dif f er entia bility.
When a SINGLE-VALUED and CONTINUOUS function is NOT differentiable at a
particular point or instant, it may have more than one tangent at that instant. In
example 1 on page 87, f(x) is SINGLE-VALUED and CONTINUOUS. But f(x) is NOT
differentiable at x = a. At the co-ordinates ( a, f(a) ) we may draw infinitely many
102 We may relate the Analysis view of differentiable and the Analytical geometr y
property of the tangent and say :
f(x) is DIFFERENTIABLE at a ⇔ f(x) has a UNIQUE TANGENT at f(a). Suppose we know the tangents z1(x), z2(x), z3(x), . . . at x1, x2, x3, . . . respectively,
can we find the curve...
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- Fall '09