Differentiation: past paper questions & worked solutions

For a positive gradient,

Since the quadratic opens upwards,

Differentiate:

At the minimum point, the gradient is 0.

Find the \(y\)-coordinate:

The tangent at a minimum point is horizontal, so its equation is

The gradient is \(\frac13\), so \[ 12x^2-8=\frac13 \]

The acceleration is \(5\), so

Differentiate:

So

Apply the power rule to each term of y=x^3+5x^2-8x.

At a stationary point the derivative is zero.

Set each factor equal to zero.

P has the smaller x-coordinate, so \(x_P=-4\). Then \[y_P=(-4)^3+5(-4)^2-8(-4)=-64+80+32=48.\]

Choose the left-hand stationary point and substitute into the original curve equation.

Multiply the three dimensions and require them all to be positive.

Differentiate each power of x.

At a stationary point, \(9x(9-2x)=0\), giving \(x=0\) or \(x=4.5\). Only \(x=4.5\) is inside the domain.

A stationary volume has zero derivative.

The derivative is positive for \(0<x<4.5\) and negative for \(4.5<x<6.75\). Hence \(x=4.5\) gives the maximum.

The volume increases then decreases; its limiting endpoint volumes are zero.

Substitute the maximising value into the original volume.

So

Differentiate:

At a maximum,

Now substitute into the volume:

The gradient is \(8\), so \[ 3x^2-40=8 \]

When \(x=4\): \[ y=4^3-40(4)+1=-95 \]

When \(x=-4\): \[ y=(-4)^3-40(-4)+1=97 \]

Point \(A\) has the greater \(x\)-coordinate, so \(x=6\).

When \(v=0\), \[ 8t-\frac{125}{t^2}=0 \]

Now find \(s\): \[ s=4(2.5)^2+\frac{125}{2.5} \]

The direction of motion reverses when velocity is zero:

Since \(t\geq 0\), \(t=\frac32\).

At \(t=\frac32\),

Differentiate:

At the stationary point, \(x=\frac23\), so the gradient is 0:

Now find \(a\), the \(y\)-coordinate when \(x=\frac23\):

Volume of the solid is cylinder plus hemisphere: \[ V=\pi x^2(20-4x)+\frac23\pi x^3 \]

So \[ y=60x^2-10x^3 \]

Differentiate: \[ \frac{dy}{dx}=120x-30x^2 \]

For a maximum, \[ 120x-30x^2=0 \]

So the valid value is \[ x=4 \]

At rest, \[ 3t^2+10t-8=0 \]

Factorise: \[ (3t-2)(t+4)=0 \]

So \[ t=\frac23 \quad \text{or} \quad t=-4 \]

Since \(t\ge0\), \[ t=\frac23 \]

Now find \(s\): \[ s=\left(\frac23\right)^3+5\left(\frac23\right)^2-8\left(\frac23\right)+10 \]

Velocity is the rate of change of displacement:

Complete the square:

Since \((t-3)^2\ge 0\), the minimum value of \(v\) is

The velocity is always positive, so the minimum speed is also \(6\).

Differentiate to find velocity:

Set \(v=0\):

Now find \(s\):

$$\\frac{dy}{dx} = 3x^2 - 16$$

3*x*² − 16 = 11 3*x*² = 27 *x*² = 9 ⟹ *x* = 3 or *x* =  − 3

For *x* = 3: *y* = (3)³ − 16(3) + 7 = 27 − 48 + 7 =  − 14

For *x* =  − 3: *y* = (−3)³ − 16(−3) + 7 =  − 27 + 48 + 7 = 28

**Step 4: Find the y-intercept (*b*) for each tangent (*y* = 11*x* + *b*).**

Using (3,−14):  − 14 = 11(3) + *b* ⟹  − 14 = 33 + *b* ⟹ *b* =  − 47

Using (−3,28): 28 = 11(−3) + *b* ⟹ 28 =  − 33 + *b* ⟹ *b* = 61

Differentiate:

The gradient of the tangent is \(11\), so

When \(x=3\):

The tangent has equation \(y=11x+b\), so

When \(x=-3\):

At a turning point, \[ 12x^2+10x+2=0 \]

When \(x=-\frac13\), \[ y=-\frac7{27} \]

When \(x=-\frac12\), \[ y=-\frac14 \]

(b) The gradient is \(4\), so \[ 12x^2+2x-20=4 \]

Factorise: \[ (3x-4)(2x+3)=0 \]

So \[ x=\frac43 \quad \text{or} \quad x=-\frac32 \]

(b) The curve has a positive gradient when

To 3 significant figures, the roots are

Since the quadratic opens upwards, it is positive outside the roots.

Particle \(P\):

Particle \(Q\):

For \(t>1\):

- \(P\) moves in the negative direction when \(1<t<\frac53\). - \(P\) moves in the positive direction when \(t>\frac53\). - \(Q\) moves in the negative direction when \(1<t<2\). - \(Q\) moves in the positive direction when \(t>2\).

So both particles move in the same direction when both velocities have the same sign.

Point \(A=(2,-6)\) lies on the curve:

Differentiate:

The gradient at \(x=2\) is 16:

Now solve

Then

For \(x=3\),

Velocity is the derivative of displacement:

At time \(T\), the velocity is \(V\), so

Use the quadratic formula:

Since \(T\ge 0\), take the positive root:

The gradient is 2, so \[ 15x^2-2x-6=2 \]

For part (a),

At the minimum point, \(x=-3\) and the gradient is 0.

For part (b),

At a horizontal tangent,

Now substitute \(x=-2\) into the curve:

First expand: \[ (2x+5)(x+1)=2x^2+7x+5 \]

For the maximum volume, \[ \frac{dV}{dx}=0 \]

Using the quadratic formula, \[ x=\frac{-1\pm\sqrt{97}}{6} \]