GATE CS 2015 Set 3 Q.60 (CS Q.50): Calculus

The velocity $v$ (in kilometer/minute) of a motorbike which starts from rest, is given at fixed intervals of time $t$ (in minutes) as follows:

t	2	4	6	8	10	12	14	16	18	20
v	10	18	25	29	32	20	11	5	2	0

The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson's $1/3^{\text{rd}}$ rule is _____.

Your answer

Official answer and solution

Accepted answers: 308 to 310
From: Official GATE 2015 answer key

Solution

Apply Simpson's 1/3 rule with step size h = 2 to compute distance as the integral of velocity.

The motorbike starts from rest, so at $t = 0$ , $v(0) = 0$ .
The time data points are given at interval $h = 2\text{ minutes}$ from $t = 0$ to $t = 20$ with $n = 10$ subintervals:

$y_0 = v(0) = 0$
$y_1 = v(2) = 10$
$y_2 = v(4) = 18$
$y_3 = v(6) = 25$
$y_4 = v(8) = 29$
$y_5 = v(10) = 32$
$y_6 = v(12) = 20$
$y_7 = v(14) = 11$
$y_8 = v(16) = 5$
$y_9 = v(18) = 2$
$y_{10} = v(20) = 0$

Simpson's $1/3^{\text{rd}}$ rule formula is:

I = \frac{h}{3} \left[ (y_0 + y_n) + 4 \sum_{i \text{ odd}} y_i + 2 \sum_{i \text{ even}} y_i \right]

Compute components:

$y_0 + y_{10} = 0 + 0 = 0$
Sum of odd ordinates: $y_1 + y_3 + y_5 + y_7 + y_9 = 10 + 25 + 32 + 11 + 2 = 80$
$4 \times 80 = 320$
Sum of even internal ordinates: $y_2 + y_4 + y_6 + y_8 = 18 + 29 + 20 + 5 = 72$
$2 \times 72 = 144$

Substitute into the formula:

I = \frac{2}{3} [0 + 320 + 144] = \frac{2}{3} \times 464 = \frac{928}{3} \approx 309.33\text{ km}

The official answer key accepts values in the range 308 to 310.

Final answer

308 to 310

Common trap

Omitting the initial point (t = 0, v = 0) stated by "starts from rest", which would throw off the number of intervals and odd/even indexing.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 3
Question: 60 of 65, printed as CS Q.50
Section: Computer Science and Information Technology

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