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Question

Given that y varies jointly with x and with the square root of z, and that y=2 when x=1/8 and z=1/4 find i.a formula giving y in terms of x and z, and ii. the value of y when whn x=3/8 and z=1/9.

Answer 1

Given that y varies jointly with x and with the square root of z, and that y=2 when x=1/8 and z=1/4 , so find i. a formula giving y in terms of x and z, and ii. the value of y when x=3/8 and z=1/9.

 

1. When we are told the y varies jointly with two things then that means y is proportional to the product of these two things, i.e. y is some constant times that product.

So, y = k*x*sqrt(z), or

y = k.x.sqrt(z)

Initial conditions

i.e. y=2 when x=1/8 and z = 1/4, so

2 = k.(1/8).sqrt(1/4)

2 = k.(1/8).(1/2) = k.(1/16)

Therefore, k = 2*16 = 32

Then y = 32.x.sqrt(z)

1. The value of y when x=3/8 and z=1/9.

We have y = 32.x.sqrt(z).

Substituting for x = 3/8 and z = 1/9,

y = 32.(3/8).sqrt(1/9)

y = 32.(3/8).(1/3)

y = 32.(1/8)

Answer: y = 4