A curve is such that d^2 y / d x^2 = 6x - 2. The gradient...?

Question

A curve is such that d^2 y / d x^2 = 6x - 2. The gradient of the curve at the point (2, -9) is 3.

(i) Express y in terms of x.

(ii) Show that the gradient of the curve is never less than -16/3

Answer 1

Integrate the given eqn

So dy/dx = 3x^2 -2x +c.

Now for x =2, dy/dx = 3 so c = - 5.

Hence dy/dx = 3x^2 - 2x - 5.

integrate again, so y = x^3 - x^2 -5x +c'.

now curve passes thr' (2, - 9)

so - 9 = 8 - 4 - 10 + c' which means c' = - 3

So y = x^2 - x^2 - 5x -3.

we must prove 3x^2 - 2x - 5 >= - 16/3 in the second part.

3x^2 - 2x - 5

= 3[ x^2 - (2/3) x] - 5

= 3[x - (1/3)]^2 - 5 - 1/3 >= - 16/3 because the sq cannot be negative.

Answer 2

d^2 y / d x^2 = 6x - 2

=> d/dx (dy/dx) = 6x - 2

Integrtaion wrt x gives

dy/dx = 3x^2 - 2x

y = x^3 - x^2

Answer 3

(i)

 6x – 2 dx

M = 3x² - 2x + c

M=3 at (2,-9)

3 = 3(2)² - 2(2) + c

c = -5

therefore dy/dx = 3x² - 2x – 5

3x² - 2x – 5 dx = x³ - x² - 5x + c

y = x³ - x² - 5x + c

(2,-9)

-9 = (2)³ - (2)² - 5(2) + c

c = -3

y = x³ - x² - 5x - 3

(ii)

Since dy/dx = 3x² - 2x – 5 <finding the minimum

using calculus method, d^2 y / d x^2 = 6x - 2

6x - 2 = 0 (stationary point)

x = 1/3 (min pt since the qradratic function has +ve x^(2) coefficient)

dy/dx (when x=1/3) = 3(1/3)² - 2(1/3) – 5

= (-1/3) - 15/3

= -16/3

therefore, the minimum gradient of the curve = -16/3

thus it cannot be less than -16/3

for the (ii) approach, u can use graph to find the min value!

hope this help

Answer 4

I think when you say 'gradient' you mean 'first derivative' otherwise this question does not make any sense. The gradient is a vector, so the gradient cannot be 3.

For (i), first integrate d^2y/dx^2 once to get that

dy/dx = 3x^2 - 2x + C.

Then the derivative at (2,-9) being 3 means that if you plug 2 in for x in dy/dx, you'll get 3. So we'll set up that equation:

3 = dy/dx(2) = 3(2)^2 - 2(2) + C.

Solve to find C = -5.

Now integrate dy/dx (using C = -5) to find that

y = x^3 - x^2 - 5x + D

Since the point (2,-9) is on this curve, we can set y = -9, x = 2 and solve for D to find that D = -3.

So we have: y = x^3 - x^2 -5x - 3

To do (ii), just use the first to show that dy/dx has local minimum at the point x = 1/3. How do we do this? Look for critical points of the second derivative, so solve for x in

6x - 2 = 0

To find that x = 1/3. We know this corresponds to a local minimum and hence global minimum because dy/dx is a parabola pointing upwards which only has one critical point. You plug x = 1/3 into the function dy/dx to see that the global minimum of dy/dx is -16/3, hence the derivative of the curve is never less than -16/3.

Hope this helps.

Answer 5

d^2 y / d x^2 = 6x - 2

==>

dy/dx=3x^2-2x+A

==>

y=x^3-x^2+Ax+B

so

3*2^2-2*2+A=3,

2^3-2^2+2A+B=-9

==>

A=-5,B=-3

==>

y=x^3-x^2-5x-3,

dy/dx=3x^2-2x-5

=3(x-1/3)^2-16/3

>=-16/3

Answer 6

d²y/dx² = 6x - 2

Integrate it to get dy/dx:

dy/dx = 3x² - 2x

Integrate it again to get y:

y = x³ - x²