Question

the graph below shows the relationship between the volume in ml, v, and the height in cm, h, for cylinders a and b. which cylinder has a greater rate of change in the height of the water, for a given change in volume?
height in cm
volume in ml

Explanation:

Step1: Recall the slope formula

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). For the relationship between volume \( V \) (x - axis) and height \( h \) (y - axis), the rate of change of height with respect to volume is the slope of the line. We can also think of the rate of change of volume with respect to height as the reciprocal of the slope (since if \( h = mV + b \), then \( V=\frac{1}{m}h-\frac{b}{m} \), so the rate of change of \( V \) with respect to \( h \) is \( \frac{1}{m} \) when \( b = 0 \) as both lines pass through the origin).

First, let's find the slope for cylinder B. We have two points for cylinder B: \((10, 0.6)\) and \((30, 1.8)\).

Step2: Calculate the slope for cylinder B

Using the slope formula \( m_B=\frac{h_2 - h_1}{V_2 - V_1} \), where \( (V_1, h_1)=(10, 0.6) \) and \( (V_2, h_2)=(30, 1.8) \)
\( m_B=\frac{1.8 - 0.6}{30 - 10}=\frac{1.2}{20}=0.06 \)

Now, let's find the slope for cylinder A. Let's assume a point on cylinder A. Let's take the point where \( V = 100 \) (we can see from the graph that the line for A goes towards \( (100, 2) \) approximately, since when \( h = 2 \), \( V = 100 \) as the line passes through the origin \((0,0)\) and \((100, 2)\)). So for cylinder A, \( (V_1, h_1)=(0, 0) \) and \( (V_2, h_2)=(100, 2) \)

Step3: Calculate the slope for cylinder A

Using the slope formula \( m_A=\frac{h_2 - h_1}{V_2 - V_1} \)
\( m_A=\frac{2 - 0}{100 - 0}=\frac{2}{100}=0.02 \)

Now, the rate of change of volume with respect to height is the reciprocal of the slope (since \( V=\frac{1}{m}h \) when \( h = mV \)). So for cylinder B, the rate of change of volume with respect to height is \( \frac{1}{m_B}=\frac{1}{0.06}=\frac{100}{6}\approx16.67 \) (mL per cm). For cylinder A, the rate of change of volume with respect to height is \( \frac{1}{m_A}=\frac{1}{0.02} = 50 \) (mL per cm)? Wait, no, wait. Wait, actually, the rate of change of volume with respect to height: if \( h = mV \), then \( V=\frac{1}{m}h \), so the derivative \( \frac{dV}{dh}=\frac{1}{m} \). But we can also think of the rate of change of height with respect to volume as \( m=\frac{dh}{dV} \), so the rate of change of volume with respect to height is \( \frac{dV}{dh}=\frac{1}{m} \). But maybe a better way: the rate of change of volume with respect to height is how much volume changes per unit height. So for cylinder B, when \( h \) changes from 0.6 to 1.8 (a change of \( \Delta h=1.8 - 0.6 = 1.2 \) cm), the volume changes from 10 to 30 (a change of \( \Delta V = 30 - 10=20 \) mL). So \( \frac{\Delta V}{\Delta h}=\frac{20}{1.2}=\frac{50}{3}\approx16.67 \) mL per cm. For cylinder A, when \( h = 2 \) cm, \( V = 100 \) mL, so \( \frac{\Delta V}{\Delta h}=\frac{100}{2}=50 \) mL per cm? Wait, no, that's not right. Wait, no, the slope of the line \( h \) vs \( V \) is \( \frac{dh}{dV} \). So for cylinder B, \( \frac{dh}{dV}=\frac{1.8 - 0.6}{30 - 10}=\frac{1.2}{20}=0.06 \) cm per mL. So the rate of change of volume with respect to height is \( \frac{dV}{dh}=\frac{1}{0.06}=\frac{50}{3}\approx16.67 \) mL per cm. For cylinder A, \( \frac{dh}{dV}=\frac{2 - 0}{100 - 0}=0.02 \) cm per mL, so \( \frac{dV}{dh}=\frac{1}{0.02}=50 \) mL per cm. Wait, but that would mean A has a higher rate? But that contradicts. Wait, maybe I mixed up. Let's re - express. The question is "Which cylinder has a greater rate of change in the height of the water, for a given change in volume?" Wait, no, the question is "Which cylinder has a greater rate of change in the height of the water, for a given change in volume?" Wait, the rate of change of height with respect to volume is \( \frac{\Delta h}{\Delta V} \), which is the slope of the \( h \) vs \( V \) graph. So for cylinder B, between \( V = 10 \) and \( V = 30 \) ( \( \Delta V=20 \) ), \( \Delta h = 1.8 - 0.6 = 1.2 \), so \( \frac{\Delta h}{\Delta V}=\frac{1.2}{20}=0.06 \) cm per mL. For cylinder A, between \( V = 0 \) and \( V = 100 \) ( \( \Delta V = 100 \) ), \( \Delta h=2 - 0 = 2 \), so \( \frac{\Delta h}{\Delta V}=\frac{2}{100}=0.02 \) cm per mL. Since \( 0.06>0.02 \), cylinder B has a greater rate of change of height with respect to volume. But the question is "Which cylinder has a greater rate of change in the height of the water, for a given change in volume?" So the rate of change of height with respect to volume is \( \frac{dh}{dV} \), which is the slope. So since the slope of B (0.06) is greater than the slope of A (0.02), cylinder B has a greater rate of change of height with respect to volume.

Wait, maybe the question is "Which cylinder has a greater rate of change in volume for a given change in height?" Then we need to calculate \( \frac{\Delta V}{\Delta h} \). For cylinder B: take two points \((10, 0.6)\) and \((30, 1.8)\). \( \Delta V=30 - 10 = 20 \), \( \Delta h=1.8 - 0.6 = 1.2 \), so \( \frac{\Delta V}{\Delta h}=\frac{20}{1.2}=\frac{50}{3}\approx16.67 \) mL per cm. For cylinder A: take two points \((0, 0)\) and \((100, 2)\). \( \Delta V = 100-0 = 100 \), \( \Delta h=2 - 0 = 2 \), so \( \frac{\Delta V}{\Delta h}=\frac{100}{2}=50 \) mL per cm. So in this case, cylinder A has a greater rate of change of volume with respect to height.

But the original question is "Which cylinder has a greater rate of change in the height of the water, for a given change in volume?" So rate of change of height with respect to volume is \( \frac{\Delta h}{\Delta V} \). For cylinder B: \( \frac{1.2}{20}=0.06 \) cm per mL. For cylinder A: \( \frac{2}{100}=0.02 \) cm per mL. So 0.06 > 0.02, so cylinder B has a greater rate of change of height with respect to volume.

Let's confirm with the slope formula. For cylinder B, using points \((10, 0.6)\) and \((30, 1.8)\):

Slope \( m_B=\frac{1.8 - 0.6}{30 - 10}=\frac{1.2}{20}=0.06 \)

For cylinder A, using points \((0, 0)\) and \((100, 2)\):

Slope \( m_A=\frac{2 - 0}{100 - 0}=0.02 \)

Since \( m_B>m_A \), the rate of change of height with respect to volume (slope) is greater for cylinder B. So cylinder B has a greater rate of change in the height of the water for a given change in volume.

Answer:

Cylinder B