Differentiation Short Questions (Question 22 – 25)


Question 22:
Given that   y = 3 4 x 2 , find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:
y = 3 4 x 2 d y d x = ( 2 ) 3 4 x = 3 2 x δ y = 47.7 48 = 0.3 Approximate change in  x  to  y δ x δ y d x d y δ x = d x d y × δ y δ x = 2 3 x × ( 0.3 ) δ x = 2 3 ( 8 ) × ( 0.3 ) y = 48 3 4 x 2 = 48 x 2 = 64 x = 8 δ x = 0.025



Question 23:
Given that y = 15 x + 24 x 3 ,

(a) Find the value of d y d x when x = 2,
(b) Express in terms of k, the approximate change in ywhen x changes from 2 to
      2 + k, where k is a small change.

Solution:
(a)
y = 15 x + 24 x 3 y = 15 x + 24 x 3 d y d x = 15 72 x 4 d y d x = 15 72 x 4 When  x = 2 d y d x = 15 72 2 4 = 10.5
(b)
Approximate change in  y  to  x  in terms of  k , δ y δ x d y d x δ y = d y d x × δ x δ y = 10.5 × ( 2 + k 2 ) δ y = 10.5 k


Question 24:
If the radius of a circle increases from 4 cm to 4.01 cm, find the approximate increase in the area.

Solution:
Area of circle,  A = π r 2 d A d r = 2 π r Approximate increase in the area to radius, δ A δ r d A d r δ A = d A d r × δ r δ A = ( 2 π r ) × ( 4.01 4 ) δ A = [ 2 π ( 4 ) ] × ( 0.01 ) δ A = 0.08 π  cm 2



Example 25:
Given that y =3t+ 5t2 and x = 5t 1.
(a) Find d y d x  in terms of x,
(b) If xincreases from 5 to 5.01, find the small increase in t.

Solution:
y = 3 t + 5 t 2 d y d t = 3 + 10 t x = 5 t 1 d x d t = 5
(a)
d y d x = d y d t × d t d x d y d x = ( 3 + 10 t ) × 1 5 d y d x = 3 + 10 ( x + 1 5 ) 5 x = 5 t 1 t = x + 1 5 d y d x = 3 + 2 x + 2 5 d y d x = 5 + 2 x 5
(b)
Small increase in  t  to  x , δ t = d t d x × δ x δ t = 1 5 × ( 5.01 5 ) δ t = 0.002

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