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 dy dx =( 2 ) 3 4 x= 3 2 x δy=47.748=0.3 Approximate change in x to y δx δy dx dy δx= dx dy ×δy δx= 2 3x ×( 0.3 ) δx= 2 3( 8 ) ×( 0.3 ) y=48 3 4 x 2 =48 x 2 =64 x=8 δx=0.025


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

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

Solution:
(a)
y=15x+ 24 x 3 y=15x+24 x 3 dy dx =1572 x 4 dy dx =15 72 x 4 When x=2 dy dx =15 72 2 4 =10.5
(b)
Approximate change in y to x in terms of k, δy δx dy dx δy= dy dx ×δx δy=10.5×( 2+k2 ) δy=10.5k


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 dA dr =2πr Approximate increase in the area to radius, δA δr dA dr δA= dA dr ×δr δA=( 2πr )×( 4.014 ) δA=[ 2π( 4 ) ]×( 0.01 ) δA=0.08π  cm 2


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

Solution:
y=3t+5 t 2 dy dt =3+10t x=5t1 dx dt =5
(a)
dy dx = dy dt × dt dx dy dx =( 3+10t )× 1 5 dy dx = 3+10( x+1 5 ) 5 x=5t1 t= x+1 5 dy dx = 3+2x+2 5 dy dx = 5+2x 5
(b)
Small increase in t to x, δt= dt dx ×δx δt= 1 5 ×( 5.015 ) δt=0.002