Tap Temperature Linear Relationship Calculation

The problem states that the water temperature ($T$) from the tap depends linearly on the proportion of the tap opening ($x$). This can be represented by the equation:

$T = mx + c$

where $m$ is the slope and $c$ is the y-intercept.

Determining the Linear Equation

We are given two points:

• When the tap is half-open ($x = \frac{1}{2}$), the temperature is $T = 40$°C. $40 = m \left( \frac{1}{2} \right) + c \quad (1)$
• When the tap is three-fourths open ($x = \frac{3}{4}$), the temperature is $T = 30$°C. $30 = m \left( \frac{3}{4} \right) + c \quad (2)$

Subtract equation (1) from equation (2) to find $m$:

$ (30 - 40) = \left( \frac{3}{4}m + c \right) - \left( \frac{1}{2}m + c \right) $

$ -10 = \frac{3}{4}m - \frac{1}{2}m $

$ -10 = \frac{1}{4}m $

$ m = -40 $

Substitute $m = -40$ into equation (1) to find $c$:

$ 40 = (-40) \left( \frac{1}{2} \right) + c $

$ 40 = -20 + c $

$ c = 60 $

The linear equation is:

$ T = -40x + 60 $

Calculating Tap Opening for 50°C

We need to find the tap opening ($x$) when the temperature ($T$) is 50°C:

$ 50 = -40x + 60 $

$ 40x = 60 - 50 $

$ 40x = 10 $

$ x = \frac{10}{40} = \frac{1}{4} $

Therefore, the tap should be $\frac{1}{4}$ (one-fourth) open to get water at 50°C.