Given that p=(2q^2-5)^2. when q=3, it is increased by 0.7%. find the appropriate percentage in p.

Question

solution
dp = 2(2q^2-5) (2dq)
dp = (4x(18-5)(0.7%)
dp = 4(13)(0.7%)

am lost from here

is dp=52x0.7/100= 0.364%

Responses

Differential Calculus - bobpursley, Friday, August 14, 2009 at 1:52pm
p=(2q^2-5)^2

dp=2(2q^2-5)(4q)dq

ok from here

dp= -8

Answer 1

you have the expression for dp.

q=3, dq=.007*3

Now solve for dp and solve for P at q=3

percentage change in p then is dp/p divided by 100.

Answer 2

y^4=c(x^2+4y^2)