NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Gradient as a measure or rate

Lesson

Imagine a fictitious experiment in which a nutrient in a range of ten amounts is given to ten identical plant specimens to find out how the growth rate depends on the supply of nutrient. The nutrient is measured in amounts of $1,2,3,...,10$1,2,3,...,10 units and each plant is given its allotted dose on each day of the experiment.

In this imaginary experiment, it is found that within certain limits the rate of growth $g$`g` in millimetres per day is $g=2+0.25u$`g`=2+0.25`u` where $u$`u` is the number of units of nutrient given to a plant each day. This is interpreted by the experimenter to mean that the plants grow at a base rate of $2$2 mm per day and this rate is increased by an amount proportional to the quantity of nutrient supplied. The additional daily amount of growth is $0.25$0.25 mm per unit of nutrient. This number is both a *rate* and the gradient of the line $g=2+0.25u$`g`=2+0.25`u`.

For plant number five, the rate of growth must have been $3.25$3.25 mm per day. We could draw a graph of the cumulative amounts of growth, day-by-day, for this plant if we knew its initial size.

Suppose plant number five was $5$5 mm high at the beginning of the experiment. After $1$1 day it must have measured $8.25$8.25 mm. After $2$2 days it would have been $11.5$11.5 mm. After $3$3 days, $14.75$14.75 mm and after $4$4 days, $18$18 mm. The increase each day is $3.25$3.25 mm so that the graph of the height of the plant after $d$`d` days must be a line with gradient $3.25$3.25. In fact, the equation of the line is $y=3.25d+5$`y`=3.25`d`+5 where $y$`y` is the height of the plant after $d$`d` days.

The number $3.25$3.25 is both the daily rate of increase of the height of the plant and it is the gradient of the graph.

Often, the graph showing the levels of a quantity as another quantity is varied will not be a line. Consider the speed-time graph from another chapter.

Suppose the graph represents the motion of a car. The speed is varying, so the occupants of the car would feel a sense of acceleration.

We measure acceleration as the change in speed per unit of time, or in other words, the time rate of change of speed. That is, we form the ratio $\frac{\delta v}{\delta t}$`δ``v``δ``t` of the change in speed over the length of the time interval. This is really the *average *acceleration over the time interval $\delta t$`δ``t` but it becomes the instantaneous acceleration when we let the size of the interval $\delta t$`δ``t` approach zero. The ratio $\frac{\delta v}{\delta t}$`δ``v``δ``t` approaches a definite limit as $\delta t\rightarrow0$`δ``t`→0 and this limit is what we mean by the instantaneous rate of change of the speed at time $t$`t`.

But, from another point of view, the gradient of the speed-time graph at time $t$`t` is just $\lim_{\delta t\rightarrow0}\frac{\delta v}{\delta t}$lim`δ``t`→0`δ``v``δ``t`. So, again, we see the equivalence of gradient and rate.

A plane starts at an altitude of $0$0 metres and ascends $160$160 metres each minute until it reaches cruising altitude.

Complete the table of values.

Time after take-off ($x$ `x`minutes)$0$0 $1$1 $2$2 $3$3 $10$10 Altitude ($y$ `y`metres)$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ State the equation relating altitude ($y$

`y`) and time after take-off ($x$`x`).Graph the equation $y=160x$

`y`=160`x`Loading Graph...State the gradient of the straight line.

What does the gradient of the straight line represent?

The rate at which the plane is gaining altitude.

AThe altitude of the plane $x$

`x`minutes after take-off.BThe rate at which the plane is gaining altitude.

AThe altitude of the plane $x$

`x`minutes after take-off.B

A landscaper is planning a garden design for an upcoming garden show, and is planning to create the garden in the shape of a rectangle. She has a total area of $320$320 m^{2} to work with.

The width $x$`x` metres and length $y$`y` metres of her design are related by the equation $xy=320$`x``y`=320, where $x>0$`x`>0. The graph of $xy=320$`x``y`=320 is given below.

Loading Graph...

Why is the rate of change of $y$

`y`always negative for $x>0$`x`>0?As the width $x$

`x`of the garden increases, the length $y$`y`increases.AAs the width $x$

`x`of the garden increases, the length $y$`y`decreases.BAs the width $x$

`x`of the garden increases, the length $y$`y`increases.AAs the width $x$

`x`of the garden increases, the length $y$`y`decreases.BThe tangent line at $x=8$

`x`=8 has been graphed below.Loading Graph...Find the instantaneous rate of change of $y$

`y`at $x=8$`x`=8.We found that at $x=8$

`x`=8, the instantaneous rate of change is $-5$−5. What does this represent?If the landscaper chooses a width of $8$8 metres, the length of the garden will be $5$5 metres.

AOf all the possible widths, the average width is $5$5 metres.

BWhen the width of the garden is $8$8 metres and increasing, the length of the garden is decreasing by $5$5 metres for each metre of change in the width.

CWhen the width of the garden is $8$8 metres, the length of the garden is $5$5 metres less.

DIf the landscaper chooses a width of $8$8 metres, the length of the garden will be $5$5 metres.

AOf all the possible widths, the average width is $5$5 metres.

BWhen the width of the garden is $8$8 metres and increasing, the length of the garden is decreasing by $5$5 metres for each metre of change in the width.

CWhen the width of the garden is $8$8 metres, the length of the garden is $5$5 metres less.

D

An object begins at a fixed point and starts moving away from it. Its distance $y$`y` metres from the fixed point after $x$`x` minutes is given by $y=x^2$`y`=`x`2, and has been graphed below.

Loading Graph...

What is the gradient of the chord joining $\left(1,1\right)$(1,1) and $\left(2,4\right)$(2,4)?

What does the gradient found in part (a) represent?

The object’s instantaneous speed at $x=2$

`x`=2 minutes.AThe object’s instantaneous speed at $x=1$

`x`=1 minute.BThe object’s average speed over a $1$1 minute interval.

CThe object’s instantaneous speed at $x=2$

`x`=2 minutes.AThe object’s instantaneous speed at $x=1$

`x`=1 minute.BThe object’s average speed over a $1$1 minute interval.

CWhat is the gradient of the chord joining $\left(1,1\right)$(1,1) and $\left(1.5,2.25\right)$(1.5,2.25)?

What does the gradient found in part (c) represent?

The object’s instantaneous speed at $t=1.5$

`t`=1.5 minutes.AThe object’s average speed over a $1$1 minute interval.

BThe object’s average speed over a $0.5$0.5 minute interval.

CThe object’s instantaneous speed at $x=1$

`x`=1 minute.DThe object’s instantaneous speed at $t=1.5$

`t`=1.5 minutes.AThe object’s average speed over a $1$1 minute interval.

BThe object’s average speed over a $0.5$0.5 minute interval.

CThe object’s instantaneous speed at $x=1$

`x`=1 minute.DFind the gradient of the tangent line at $x=1$

`x`=1.What does this gradient represent?

The object’s instantaneous speed at $t=1.5$

`t`=1.5 minutes.AThe object’s average speed over a $0.5$0.5 minute interval.

BThe object’s average speed over a $1$1 minute interval.

CThe object’s instantaneous speed at $x=1$

`x`=1 minute.DThe object’s instantaneous speed at $t=1.5$

`t`=1.5 minutes.AThe object’s average speed over a $0.5$0.5 minute interval.

BThe object’s average speed over a $1$1 minute interval.

CThe object’s instantaneous speed at $x=1$

`x`=1 minute.DAt $x=2$

`x`=2 will the instantaneous speed of the object be greater than, less than or equal to the instantaneous speed at $x=1$`x`=1?Less than

AEqual to

BGreater than

CLess than

AEqual to

BGreater than

C

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems