SOLVE THE PROBLEM

Question

The curve C has equation y=x²-Inx,x ˃ 0.Show that C has just one turning point and find its coordinates.

When the depth of liquid in a container is xcm,the volume is x(x² + 25)cm³.Liquid is added to the container at  a constant rate of 2cm³s¯¹.Find the rate of change of the depth of the liquid at the instant when x=1.

Answer 1

Question: The curve C has equation y=x²-Inx,x ˃ 0.Show that C has just one turning point and find its coordinates.

y = x² - Inx,x ˃ 0.

dy/dx = 2x - 1/x

There is a TP when y' = 0.

Setting 2x - 1/x = 0

2x² - 1 = 0

2x² = 1

x² = 1/2

x = ± 1/4

But x > 0, => x = 1/4 is the only solution

Then, y = (1/4)² - ln(1/4) = 1/16 + ln(4)

Turning point is at: (x,y) = (1/4, 1/16 + ln(4))

Answer 2

Question: When the depth of liquid in a container is xcm,the volume is x(x² + 25)cm³.Liquid is added to the container at  a constant rate of 2cm³s¯¹.Find the rate of change of the depth of the liquid at the instant when x=1.

V = x(x² + 25)cm³, x >= 0

dV/dt = 2 cm³s¯¹

Rate of change of depth is dx/dt, where

dx/dt = (dx/dV).(dV/dt)

Now, dV/dx = 3x² + 25, so dx/dV = 1/(3x² + 25)

And, dx/dt = (1/(3x² + 25)).(2)

dx/dt = 2/(3x² + 25)

at x = 1,

dx/dt = 2/(3 + 25) = 2/28

dx/dt = 1/14 cm³s¯¹