Geometrical application of Calculus (1 Viewer)

[roosterman]

CAn someone please help!

1)y=ax^3+ bx^2 + cx + d has a local maximum at the point (-2, 27) and a local minimum at (1,0)

find the values of a, b, c and d.

2) Find the values of x for which y''= 0 if y=x(x+1)^1/2

3) A rectangle has a constant area of 36 cm^2.

a) if the length of the rectangle is x, show that its perimeter is

P = 2x + 72/x

b) show that dP/dx= 2(x - 6)( x + 6)/x^2.

 

[Templar]

1. y=ax^3+bx^2+cx+d
27=-8a+4b-2c+d
0=a+b+c+d

y'=3ax^2+2bx+c
0=12a-4b+c
0=3a+2b+c

Solving simultaneously, a=2, b=3, c=-12, d=7

2. y=xsqrt(x+1)
y'=x/(2sqrt(1+x)+sqrt(1+x)
y"=-x/(4(1+x)^3/2)+1/sqrt(1+x)

x=-4/3

3. A=bh
36=xb
b=36/x

P=2(h+b)
=2(x+36/x)
=2x+72/x

P'=2-72/x^2
=2(x^2-36)/x^2
=2(x+6)(x-6)/x^2

 

[pLuvia]

2) Find the values of x for which y''= 0 if y=x(x+1)^1/2

y=x(x+1)^1/2
y' = x/2[sqrt(x+1)] + sqrt(x+1)
y" = Differentiate again and make it equal to 0 to solve for y"

3) A rectangle has a constant area of 36 cm^2.

a) if the length of the rectangle is x, show that its perimeter is

P = 2x + 72/x

b) show that dP/dx= 2(x - 6)( x + 6)/x^2.

a) Letting x be the length and y be the width
since xy = 36
y = 36/x
.: Perimeter = x+x+36/x+36/x = 2x+72/x
as req'd

b)P = [2x² + 72] / x
dP/dx = Using quotient rule
= [x*4x - (2x²+72)] / x²
= [4x² - 2x²-72] / x²
= [2x² - 72] / x²
= 2(x² - 36)] / x²
= 2(x-6)(x+6) / x²

 

[roosterman]

cheers thanx!

 

[roosterman]

also for f(x)= 2x^3 - 3x^2 + 5x + 1
deduce that the equation F(x)=0 has only one real root

 

[Sober]

also for f(x)= 2x^3 - 3x^2 + 5x + 1
deduce that the equation F(x)=0 has only one real root

f(x)= 2x^3 - 3x^2 + 5x + 1

f'(x) = 6x^2 - 6x + 5

Solve for f'(x)=0:

x = (6 + sqrt(-84)) / 12

No real solution therefore no turning points, if it has no turning points it must have 1 root.

 

[YBK]

No real solution therefore no turning points, if it has no turning points it must have 1 root.

we learnt that in 4unit...

 

[Sober]

we learnt that in 4unit...

really? I just figured it from commen sense...

 

[YBK]

really? I just figured it from commen sense...

yup, that's okay - teacher explaied a similar problem to me through common sense. It's just that we learnt in 4u that if the highest power is a cube and there is only one real root, then the other 2 are complex conjugates.

A cubic equation has 3 roots:

2 non-real (complex conjugate pairs) and 1 real
OR
3 real roots

(I think that's right)

 

[Sober]

Or more generally, a polynomial of n degree has n roots.

If n is odd then an odd number are real (could be either 1, 3, 5... or n), and the rest are complex.

If n is even then an even number are real (could be either 0, 2, 4...or n) and the rest are complex.

Bear in mind that these roots are not necesserily distinct.