Differentiation

can anyone help me with these questions:

a curve has the equation y= 12x^0.5 - x^1.5

a) show that dy/dx= 1.5x^-0.5(4-x)^0.5

b) find the coordinates of the point on the curve where the gradient is 0

Also...
a curve c has the equation y= x^3 -5x^2 +5x + 2
a) find dy/dx in terms of x
b) the points p and Q lie on c the gradient of c at both p and q is 2 the x coordinate of p is 3
i- find the x coordinate of Q
ii-find and equation for the tangent to c and p, giving answer in form y=mx+c where m and c are constants

Finally (sorry)
if this tangent intersects the coordinate axes at the points R and S find the length of RS giving answere as surd

Any help would be great

Reply 1

London Prophet

a)

y = 12x^(0.5) - x^(1.5)

dy/dx = 0.5*12x^(0.5 - 1) - 1.5*x^(1.5 - 1)

= 6x(-0.5) - 1.5x^(0.5)

b)

if gradient = 0 then dy/dx = 0

therefore: 6x(-0.5) - 1.5x^(0.5) = 0

think of this as: (6 / [Root(x)]) - (1.5*[Root(x)]) = 0

take (1.5*[Root(x)]) to the other side

(6 / [Root(x)]) = 1.5*[Root(x)]

square both sides

36/x = 2.25x

multiply by x

36 = 2.25x^(2)

x = [Root(36/2.25)] = 16

x = +4 or -4

Reply 2

London Prophet

jimmykite

a curve c has the equation y= x^3 -5x^2 +5x + 2
a) find dy/dx in terms of x
b) the points p and Q lie on c the gradient of c at both p and q is 2 the x coordinate of p is 3
i- find the x coordinate of Q
ii-find and equation for the tangent to c and p, giving answer in form y=mx+c where m and c are constants

Finally (sorry)
if this tangent intersects the coordinate axes at the points R and S find the length of RS giving answere as surd

Any help would be great

Remember if y = x^n
Then dy/dx = nx^(n-1)

so:

a)

dy/dx = 3*x^(3-1) - 2*5x^(2-1) +1*5x(1-1) + 0*2(0-1)
dy/dx = 3x^(2) - 10x + 5 + 0

b)i)

points p and q are at points where the gradient of c = 2

so dy/dx = 2

3x^2 - 10x + 5 = 2
bring terms to one side
3x^2 - 10x +3 = 0
factorise
(3x-1) (x-3) = 0
therefore
x = 1/3 or 3

well the x-coordinate of Q is either 1/3 or 3. On the unstated assumption that P and Q aren't the same point, then if P is when x=3, then Q is when x=(1/3)

b)ii)

I think you have written the question wrong, it would make mroe sense as "what is the tangent of c AT point P"

we know the gradient at P is 2,

so y = mx + c where m is the gradient
becomes
y = 2x + c

to work out c you need the co-ordinates of P, which are (3, y)

put into original formula

y= (3)^3 - 5*(3)^2 +5*(3) + 2
y = 27 - 45 + 15 + 2
y = -1

so: y = 2x + c at a point (3, -1)

so: -1 = 2*(3) + c

therefore: c = -(1/6)

so: y = 2x -(1/6)

Reply 3

Revelation

jimmykite

Also...
a curve c has the equation y= x^3 -5x^2 +5x + 2
a) find dy/dx in terms of x
b) the points p and Q lie on c the gradient of c at both p and q is 2 the x coordinate of p is 3
i- find the x coordinate of Q
ii-find and equation for the tangent to c and p, giving answer in form y=mx+c where m and c are constants

a) y = x^3 - 5x^2 + 5x + 2
dy/dx = 3x^2 - 10x + 5

b) gradient is 2 when the x co-ordinate of p = 3 .'.
2 = 3x^2 - 10x + 5
3x^2 - 10x + 3 = 0
x^2 - 10/3x + 1 = 0 {complete the square}
(x - 5/3)^2 - 25/9 + 1 = 0
(x-5/3)^2 = 16/9
x-5/3 = +/- 4/3
x = 5/3 +/- 4/3
either x = 3 or x = 1/3

.'. x co-ordinate of Q = 1/3

c) erm i assume finding the tangent to C at the point p?
gradient at point p, m = 2
x co-ordinate of p = 3
when x=3 => y = 3^3 -5*3^2 +5*3 + 2
= 27 - 45 + 15 + 2
= -1

y-y1 = m(x-x1)
y--1 = 2(x-3)
y+1 = 2x - 6
y = 2x - 5

meh im probably wrong