1. 14. Functions and Derivatives
Objectives:
Tangents to curves
Rates of Change -
Applications of the Chain Rule
Refs: B&Z

2. 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).

3. Step 1: Differentiate = e2x-1. 2 = 2 e2x-1. Step 2: Evaluate y = e2x-1dx dy
= e2x-1. 2
= 2 e2x-1.
This will tell us the gradient of the tangent to the curve
at any point x.
Step 2: Evaluate dx dy
at the point x=1/2 dx dy
= 2 e2x-1 so dx dy
(1/2) = 2 e2(1/2)-1 = 2e0 = 2.

4. Step 3: Remember that the equation for a straight line is
y = mx + c.
We have already calculated m=2. So the equation for our
tangent 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.

5. Rate of Change For a function y = f(x) we may interpret the quotient
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 travelled
in time t, then is the rate of change of distance with
respect to time. dy dt

6. Example A plane travels a distance of y kms in time t hours.
For any 0 ≤ t ≤ 2, the distance is calculated according to
the formula
y(t) = 800(t2-1/3 t3).
What is the velocity of the plane at time t=1?
We need to calculate the rate of change of distance with
respect to time.
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.

7. 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? x
V(volume) = x3.
The length of the side is changing with time
and so is the volume.
(rate of change of volume) when
We want to know dV dt
x= 3.0 cm.
We know that dx dt
= 0.1 cm for any t.

8. According to the chain rule
Now, V=x3  dV dx
= 3x2 and dt
= 0.1 cm.
Now applying the chain rule we have dV dt
=3x2(0.1) = 0.3 x2 .
At x=3.0 cm we have dV dt
= 0.3(3.0)2=2.7 cm3
So when x=3cm, the volume is increasing by 2.7 cm3 per day.

9. Example A 6m ladder is placed against a wall (which is perpendicular
to 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?

10. Solution: x
wall 6m y
Both x (distance from ladder to wall) and y (height of ladder
from ground) are changing with time (t). dx dt
We want to know
when x=3m.
We know dy dt
= -2/3 m/sec and by Pythagorus, 62=x2+y2.
So when x=3, y2= 27.
Also x = (36-y2)1/2.

11. Now dy dt dx = . So dx dy =
1/2(36-y2)-1/2 . (-2y)
-y(36-y2)-1/2. dx dt
and =
-y(36-y2)-1/2 . (-2/3)
=2y/3(36-y2)-1/2
(by the chain rule)
x=3, y2= 27 so
Now when
(36-27)-1/2 =2√3(9)-1/2 = dx dt
= 2(27)1/2 3
2√3
m/sec

12. Example
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 of
increase in pressure (kpa/min) at that time?
Solution dP dt
We want to know
when V = 2.4.

13. We know that dV dt
= m3/min at V=2.4
and that P=320 when V=2.4. dP dt dV
Now = .
by the chain rule
dP . dV
So when V=2.4 dP dt
= -0.05
We know that PV=constant, so when V=2.4, P=320 gives the
constant as Hence P=768V-1 and dP = -768V-2. dV
Hence dP dt =
(2.4)2
=6.7 kpa/min.

14. You should now be
able to complete
Example Sheet 5
from the Orange Book.