14. Functions and 14. Functions and DerivativesDerivatives

Objectives:1. Tangents to curves

2. Rates of Change - Applications of the Chain Rule

Refs: B&Z 10.3.

Tangents to Curves

Find the tangent to the curve y = e2x-1 at the point(x,y) = (1/2, 1).

What does this mean?

We must determine the equation of a straight line through the point (1/2, 1) which has the same slope as the curve at (1/2, 1).

Step 1: Differentiate y = e2x-1

dxdy

= e2x-1. 2 = 2 e2x-1.

This will tell us the gradient of the tangent to the curveat any point x.

Step 2: Evaluatedxdy at the point x=1/2

dxdy

(1/2) = 2 e2(1/2)-1 = 2e0 = 2.dxdy

= 2 e2x-1 so

Step 3: Remember that the equation for a straight line is y = mx + c.

We have already calculated m=2. So the equation for ourtangent is:

ytan = 2x + c.

Step 4: We now calculate c. To do this we need to know a point on the line. That’s OK since we were asked to find the tangent at (1/2,1) - so we can use this point.

ytan = 2x + c at (1/2,1) so

1 = 2(1/2) + c c = 0. So ytan = 2x.

Rate of Change

For a function y = f(x) we may interpret the quotient

€

f '(x) = limh→0

f (x+h) − f (x)

h

to be the ratio of the change in y (f(x+h)-f(x)) to the change in x (x+h-x=h).

In this way the derivative is the instantaneous rate of change(of y with respect to x).

A familiar example is velocity. If y is the distance travelledin time t, then is the rate of change of distance with

respect to time.

dydt

What is the velocity of the plane at time t=1?

ExampleA plane travels a distance of y kms in time t hours.

For any 0 ≤ t ≤ 2, the distance is calculated according tothe formula

y(t) = 800(t2-1/3 t3).

y'(t) = 800(2t-t2) for 0 ≤ t ≤ 2.

This tells us the velocity at any time 0 ≤ t ≤ 2.

So y'(1) = 800(2(1)-12) = 800 km/hr when t=1.

We need to calculate the rate of change of distance with respect to time.

Applications of the chain rule.

A cubic crystal grows so that its side length increases at a constant rate of 0.1cm per day.

When the crystal has side length 3.0 cm, at what rate is its volume increasing?

V(volume) = x3.

x

The length of the side is changing with timeand so is the volume.

We know that dxdt

= 0.1 cm for any t.

(rate of change of volume) whenWe want to know dVdtx= 3.0 cm.

According to the chain rule

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dV

dt=dV

dx.dx

dt

Now, V=x3

dVdx

= 3x2 and dxdt

= 0.1 cm.

Now applying the chain rule we have dVdt

=3x2(0.1) = 0.3 x2 .

At x=3.0 cm we havedVdt

= 0.3(3.0)2=2.7 cm3

So when x=3cm, the volume is increasing by 2.7 cm3 per day.

A 6m ladder is placed against a wall (which is perpendicularto the ground). The top of the ladder is sliding down the wall at a rate of 2/3 m/sec.

At what velocity is the bottom of the ladder moving away from the wall when the bottom of the ladder is 3m from the wall?

Example

Solution:

Both x (distance from ladder to wall) and y (height of ladder from ground) are changing with time (t).

dxdt

We want to know

when x=3m.

We know dydt

= -2/3 m/sec and by Pythagorus, 62=x2+y2.

So when x=3, y2= 27. Also x = (36-y2)1/2.

x

wall6m

y

Now

dydt

dxdy

dxdt

= .

So dxdy

=

=

1/2(36-y2)-1/2 . (-2y)

-y(36-y2)-1/2.

x=3, y2= 27 so Now when

dxdt

and = -y(36-y2)-1/2 . (-2/3)=2y/3(36-y2)-1/2

(by the chain rule)

(36-27)-1/2 =2√3(9)-1/2 = dxdt

= 2(27)1/2

3

2√33

m/sec

Gas in a large container is being compressed by an increasing load on a piston. When the volume of gas is 2.4 m3

the volume is being decreased by 0.05 m3/minute and the pressure is 320 kpa.

Assuming Boyles law (PV = constant), what is the rate ofincrease in pressure (kpa/min) at that time?

Example

SolutiondPdt

We want to know when V = 2.4.

We know that dV

dt= -0.05 m3/min at V=2.4

and that P=320 when V=2.4.

dPdt

dPdV

dVdt

Now

= . by the chain rule

dP .dV

So when V=2.4dPdt

= -0.05

We know that PV=constant, so when V=2.4, P=320 gives theconstant as 768. Hence P=768V-1 and dP = -768V-2. dVHence dP

dt= -0.05 . -768 (2.4)2

=6.7 kpa/min.

You should now beable to complete

Example Sheet 5from the Orange Book.