LOGARITHM CALCULATION IN MATHS

QUESTION

Engineering Computations One:
Differential and Integral Calculus
Assignment Three
Due Monday 7 May 2012
Please ensure that all your working is clearly set out, all pages are stapled
together, and that your name along with the course name appears on the front
page of the assignment. Please place completed assignments in the assignment
box in WT level 1 by 5pm on the due date.
Questions:
1. Find the derivative
dy
dx
for each of the following:
y = e
x cos x
x cos y + y cos x =1 y = x
y =
√
xe
x
2

x
2
+1

10
y =
x
√
x
2
ln x

1+x
=2xy.
2. On what interval is the curve y = e
−t
2
+1
2
y
2
concave downward?
3. Air is being pumped into a spherical weather balloon. At any time t,the
volume of the balloon is V (t) and its radius is r(t).
(a) What do the derivatives
dV
dr
and
dV
dt
represent?
(b) Express
dV
dt
in terms of
dr
dt
.
4. A particle is moving along the curve y =
√
x. As the particle passes
through the point (4, 2), its x-coordinate is increasing at the rate of 3
cm/s.
(a) How fast is the y-coordinate changing as it passes through the point
(4, 2)?
(b) How far is the particle from the origin as it passes through this point?
(c) How fast is the distance from the particle to the origin changing as
it passes through this point?
5. Find where the graph of the function f(x)=
x
(x−1)
has any vertical and
horizontal asymptotes, where it is increasing or decreasing, any local max-
2
imum and minimum values, and where it is concave upward or downward.
Use this information to sketch the graph of f.
1
6. Find the point on the parabola x + y
2
= 0 that is closest to the point
(0, −3).
7. A fence that is 4 metres tall runs parallel to a tall building at a distance
of 1 metre from the building. What is the length of the shortest ladder
that will reach from the ground over the fence to the wall of the building?
8. Apply Newton’s method to the equation
1
x
−a = 0 to derive the following
reciprocal algorithm:
x
n+1
=2x
n
− ax
2
n
(which enables a computer to find reciprocals without actually dividing).
Then use this algorithm to compute
2
1
1.6984
correct to five decimal places.

SOLUTION’

1. (a) Given,

 

Therefore, differentiating w. r. t. x,

 

 

(b) ) Given,

 

Therefore, differentiating w. r. t. x,

 

 

(c) Given,

 

Therefore, differentiating w. r. t. x,

 

 

(d) Given,

 

Taking logarithm on both sides, we get

 

Differentiating both sides w. r. t. x,

 

 

(e) Given,

 

Taking logarithm on both sides, we get

 

Differentiating both sides w. r. t. x,

 

(f) Given,

 

Taking logarithm on both sides, we get

 

Differentiating both sides w. r. t. x,

 

 

 

2.

Given,

 

Therefore,

 

If the curve is concave downwards, then

 

(i)             Either,

 

(ii)           Or,

 

Case (ii) is not possible.

Therefore, the interval in which the given curve is concave downwards is

 

3. (a) The derivative

 

Represents the rate of change of volume with respect to the radius of the balloon.

The derivative

 

Represents the rate of change of volume with respect to the time.

 

(b) We know that volume of the spherical balloon is,

 

Differentiating both sides with respect to t, we get

 

 

4. Given

 

At the point,

 

 

Therefore,

 

We substitute

 

Therefore,

 

Therefore, as it passes through the point (4, 2), the y-coordinate is increasing at the rate of

0.75 cm/s  Ans.

 

(b) As the particle passes through the point (4, 2), the distance of the particle from the origin

 

 

(c) Let the distance of the particle from the point P(x, y) to the origin (0, 0) be D.

Therefore,

 

 

Differentiating both sides with respect to time t, we get

 

At the point (4, 2)

 

Therefore,

 

Therefore, as the particle passes through this point, the distance from the particle to the origin is changing at the rate of 3.02 cm/sec.  Ans.

 

5. Given,

 

The domain is

 

To find the vertical asymptote,

 

 

Therefore, the vertical asymptote is

 

 

To find the horizontal asymptote,

 

 

Therefore, the horizontal asymptote is

 

i.e. x-axis.

 

To find the intervals of increase or decrease

 

 

Therefore,

 

And,

Therefore, in the interval

 

 

Therefore, the function is decreasing.

In the interval

 

 

Therefore, the function is increasing.

In the interval

 

 

Therefore, the function is decreasing.

 

To find local maxima or minima

 

 

Therefore, when

 

 

 

Therefore, the point of local minima is

 

 

To find the points of concavity

 

 

Therefore,

 

Therefore, in the interval

 

 

In the interval

 

 

In the interval

 

 

Therefore, the function is concave downwards in and concave upwards in  and.

At x=-2,

 

Therefore, the point of inflection is

 

The sketch of the curve is shown below.

 

 

 

6. Let the point be (x, y).

Let the distance of the point (x, y) from (0, -3) be D.

Therefore,

 

On the parabola,

 

Therefore,

 

Therefore, we have to minimize D. We shall minimize D². If D² is Minimum, then D is Minimum.

 

Differentiating both sides with respect to y,

 

For maxima or minima,

 

Either,

 

Or,

 

Which gives imaginary values.

Therefore,

 

Now,

 

At

 

 

Therefore, the distance is Minimum when

 

Therefore,

 

Therefore, the point closes to the point  (0, -3) is

 

 

 

 

 

 

7.

 

 

 

 

 

 

 

 

 

 

 

 

Here, AB is the wall, DE is the fence, AC is the ladder.

Therefore,

 

We suppose, that the distance between the bottom of the ladder and the fence be x.

Therefore,

 

Therefore, by Pythagoras theorem,

 

Now, triangle AFD and the triangle DEC are similar.

Therefore,

 

Therefore,

 

Differentiating both sides with respect to x,

 

For maxima or minima,

 

Either,

 

Which is not possible.

Or,

 

Now,

 

When,

 

 

Therefore, AC² is Minimum when

 

Length of the shortest ladder

 

 

8. Given,

 

Therefore,

 

If x_n is any point, then

 

 

Therefore, by Newton’s method,

 

Now, if

 

Here,

 

We take,

 

Therefore,

 

Therefore,

L030