Engineering Mathematics: Solved Assignments and Essays

Engineering Mathematics 1
ENGINEERING MATHEMATICS
Student’s Name
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Engineering Mathematics 2
Engineering Mathematics
Task #4
Part #1
Solution
Y-axis
Max turning point
X-axis
Min turning point
y = 3x2-5x
Dy/dx = y’= d (3x2-5x)/dx
d (3x2-5x)/dx = 6x-5
y’= 6x-5
At the turning point the gradient is zero thus y’ = 0
6x-5 = 0
x = 5/6
But y = 3x2-5x therefore inserting for the values of x in the equation we get the value for y
y = 3(5/6)2-5(5/6)
= 25/12 – 25/6

Engineering Mathematics 3
= 25/4
Therefore the turning point is (5/6, 25/4)
To determine whether the curve is a maxima or a minima then we differentiate it second time,
second derivative.
y’ = 6x-5 = dy/dx
d2y/dx2 = y” = d2(6x-5)/dx2
d2 (6x-5)/dx2 = 6
but y”<0 means a negative concave and y”>0 positive concave
Since the answer after the second derivative is a positive value then it means the concave curve
faces up.
Therefore the turning point is (5/6, 25/4) and the curve is a Minima.
Part #2
Find the maxima and the minima values for the function;
y = x3- 4x + 6
dy/dx = y’= d (x3- 4x + 6)/dx
D (x3- 4x + 6)/dx = 3x2-4
Dy/dx = 3x2-4
At the turning point the gradient is always zero, therefore, y’= 0
Dx/dy = 3x2-4 = 0
x2 = 4/3
x = +2/ √ 3 or -2/ √ 3
To determine the maxi and minima then we go to the second derivative;
Dy/dx = 3x2-4

Engineering Mathematics 4
d2y/dx2 = y” = d2 (3x2-4) dx2
d2y/dx2 = 6x
Y” = 6x
When x is +2/ √ 3
y” = 6(+2/ √ 3 )
= 12/√ 3 thus it is the value of the minima
When x -2/ √ 3
y” = 6(-2/ √ 3 )
= -12/√ 3 thus it is the value of the maxima
Maxima, -12/√ 3 ,
Minima, +12/ √ 3
Task #1
a) v = (t2 + 6)2
Solution
dv/dt = v’ =d(t2 + 6)2/dt
dv/dt = 2 x 2t (t2 + 6)
= 4t (t2 + 6)
But t = 5therefore,
4(5)*((5)2 + 6) = 620 seconds
= 620 seconds
b) v = (3t3-4t+6)3