Further Derivatives: exam review Exam review: second paper Worked solutions

Differentiate each function with respect to \(x\). Simplify your answers where you can. [6]

1. (a) (a) \(f(x)=\frac{10x^{4}-6x^{2}}{\sqrt{x}}\) [3]

   • M1 for splitting into separate powers: \(10x^{\frac{7}{2}} - 6x^{\frac{3}{2}}\)
   • A1 for one correct term of the derivative
   • A1 for \(35x^{\frac{5}{2}} - 9 \sqrt{x}\)

2. (b) (b) \(g(x)=3\ln\left(x^{7}\right)+5\tan x\) [3]

   • M1 for using \(\ln x^{7}=7\ln x\)
   • A1 for \(\frac{21}{x}\)
   • A1 for \(5\sec^{2}x\)

Differentiate with respect to \(x\). [6]

1. (a) (a) \(y=\sqrt[3]{5x^{2} + 7}\) [2]

   • M1 for the chain rule: \(\frac{1}{3}(5x^{2} + 7)^{- \frac{2}{3}}\) multiplied by the derivative of the bracket
   • A1 for \(\frac{10x}{3 \left(5x^{2} + 7\right)^{\frac{2}{3}}}\)

2. (b) (b) \(y=e^{\cos(2x)}\) [2]

   • M1 for \(e^{\cos(2x)}\times\) the derivative of \(\cos(2x)\)
   • A1 for \(- 2e^{\cos(2x)} \sin(2x)\)

3. (c) (c) Find the gradient of the curve \(y=(e^{x})^{2}+7x\) at the point where \(x=\ln 5\). [2]

   • M1 for \(\frac{dy}{dx}=2e^{2x}+7\)
   • A1 for \(2\times 5^{2}+7=57\)

Differentiate with respect to \(x\). Show the rule you use. [6]

1. (a) (a) \(y=x^{3}\ln(2x-1)\) [3]

   • M1 for the product rule \(uv'+vu'\)
   • A1 for \(u'=3x^{2}\) and \(v'=\frac{2}{2x-1}\)
   • A1 for \(3x^{2}\ln(2x-1)+\frac{2x^{3}}{2x-1}\)

2. (b) (b) \(y=\frac{e^{x}}{(3x-2)^{2}}\) [3]

   • M1 for the quotient rule \(\frac{vu'-uv'}{v^{2}}\)
   • A1 for \(u'=e^{x}\) and \(v'=6(3x-2)\)
   • A1 for \(\frac{e^{x}(3x-8)}{(3x-2)^{3}}\) (simplified)

The curve \(C\) has equation \(y=e^{3x}-7x\). [7]

1. (a) (a) Find \(\frac{dy}{dx}\). [1]

   • A1 for \(3e^{3x} - 7\)

2. (b) (b) The tangent to \(C\) at the point P is parallel to the line \(y=5x-1\). Find the exact \(x\)-coordinate of P. [3]

   • M1 for setting \(3e^{3x} - 7=5\)
   • A1 for \(e^{3x}=4\)
   • A1 for \(x=\frac{2 \ln(2)}{3}\)

3. (c) (c) Find the equation of the normal to \(C\) at the point where \(x=0\). Give your answer in the form \(y=mx+c\). [3]

   • A1 for gradient of the tangent \(-4\) at \(x=0\) and \(y=1\)
   • M1 for the negative reciprocal \(\frac{1}{4}\)
   • A1 for \(y=\frac{1}{4}x+1\)

The concentration of a drug in a patient's blood, in mg per litre, is modelled by \(f(t)=10te^{-0.5t}\), where \(t\) is the time in hours after the dose is taken, \(t\geq 0\). [8]

1. (a) (a) Find \(f'(t)\). [2]

   • M1 for the product rule
   • A1 for \(f'(t)=(10 - 5t)e^{-0.5t}\)

2. (b) (b) Find the exact coordinates of the stationary point of the graph of \(y=f(t)\). [2]

   • M1 for setting \(f'(t)=0\) (the exponential is never zero)
   • A1 for \(t=2\ \text{and}\ y=\frac{20}{e}\)

3. (c) (c) Find \(f''(t)\). [2]

   • M1 for differentiating \((10 - 5t)e^{-0.5t}\) using the product rule
   • A1 for \(f''(t)=(\frac{5t}{2} - 10)e^{-0.5t}\)

4. (d) (d) Use your answer to part (c) to show that the stationary point is a maximum, and find the \(t\)-coordinate of the point of inflexion. [2]

   • A1 for \(f''(2)=- \frac{5}{e}<0\), so a maximum
   • A1 for \(f''=0\) at \(t=4\) (and \(f''\) changes sign)

The light intensity, \(I\) lux, at a distance of \(x\) metres from a lamp is modelled by \(I=\frac{250}{x^{2}}\). A moth flies directly away from the lamp at a constant speed of 0.6 metres per second. [5]

1. (a) (a) Find \(\frac{dI}{dx}\). [2]

   • M1 for writing \(I=250x^{-2}\)
   • A1 for \(-\frac{500}{x^{3}}\)

2. (b) (b) Find the rate of change of \(I\) when \(x=5\). [2]

   • M1 for \(\frac{dI}{dt}=\frac{dI}{dx}\times\frac{dx}{dt}\)
   • A1 for \(-\frac{500}{5^{3}}\times 0.6=-2.40\)

3. (c) (c) Explain what the sign of your answer to part (b) tells you. [1]

   • R1 for the negative sign: the light intensity is decreasing, by 2.4 lux each second

A cyclist rides along a straight road. The displacement from a fixed point O is \(s\) metres after \(t\) seconds, where \(s=2 t^{3} - 21 t^{2} + 60 t\) for \(t\ge0\). [7]

1. (a) (a) Find expressions for the velocity and the acceleration. [2]

   • A1 for \(v=6 t^{2} - 42 t + 60\)
   • A1 for \(a=12 t - 42\)

2. (b) (b) Find the times when the particle is at rest. [2]

   • M1 for \(6 t^{2} - 42 t + 60=0\)
   • A1 for \(t=2\) and \(t=5\)

3. (c) (c) Determine whether the particle is speeding up or slowing down when \(t=6\). Justify your answer. [3]

   • M1 for substituting \(t=6\): \(v=24\), \(a=30\)
   • A1 for \(v\) positive and \(a\) positive
   • R1 for \(v\) and \(a\) having the same sign, so the particle is speeding up

A toy train runs along a straight track. The displacement from a fixed point O is \(s\) metres after \(t\) seconds, where \(s=12\left(1-e^{-0.25t}\right)\) for \(t\ge0\). [6]

1. (a) (a) Find an expression for the velocity. [2]

   • M1 for the chain rule: factor \(-1/4\) from the exponent
   • A1 for \(v=3e^{-0.25t}\)

2. (b) (b) Find the time when the velocity is \(0.9\) m s\(^{-1}\). [2]

   • M1 for \(3e^{-0.25t}=0.9\)
   • A1 for \(t=4.82\)

3. (c) (c) Find the distance from O that the particle approaches as \(t\) becomes very large. [2]

   • M1 for \(e^{-0.25t}\to0\) as \(t\to\infty\)
   • A1 for \(12\) m

A storage tank is in the shape of an inverted cone of height 15 m and radius 6 m at the top. Water is pumped into the tank at a constant rate of 8 m³ per minute. At time \(t\) minutes the depth of the water is \(h\) m and the volume of water is \(V\) m³. [12]

1. (a) (a) Show that \(V=\frac{4 \pi h^{3}}{75}\). [2]

   • M1 for similar triangles: \(\frac{r}{h}=\frac{6}{15}\), so \(r=\frac{2h}{5}\)
   • A1 for \(V=\frac{1}{3}\pi\left(\frac{2h}{5}\right)^{2}h=\frac{4 \pi h^{3}}{75}\) (AG)

2. (b) (b) Find the rate at which the depth is increasing at the instant when \(h=10\). [4]

   • M1 for \(\frac{dV}{dh}=\frac{4 \pi h^{2}}{25}\)
   • A1 for \(\frac{dV}{dh}=16 \pi\ \text{at}\ h=10\)
   • M1 for \(\frac{dh}{dt}=\frac{dV}{dt}\div\frac{dV}{dh}=8\div 16 \pi\)
   • A1 for \(\frac{1}{2 \pi}\) = \(0.159\)

3. (c) (c) Find the rate at which the area of the liquid surface is increasing at the instant when \(h=10\). [4]

   • M1 for \(A=\pi r^{2}=\frac{4 \pi h^{2}}{25}\)
   • A1 for \(\frac{dA}{dh}=\frac{8 \pi h}{25}\)
   • M1 for \(\frac{dA}{dt}=\frac{dA}{dh}\times\frac{dh}{dt}\)
   • A1 for \(\frac{8}{5}\) = \(1.60\)

4. (d) (d) Explain why the depth rises more and more slowly while the liquid flows in at a constant rate. [2]

   • R1 for \(\frac{dh}{dt}=\frac{50}{\pi h^{2}}\) decreasing as \(h\) increases
   • R1 for the reason: the surface area grows with \(h\), so the same volume spreads over a larger area

The function \(f\) is defined by \(f(x)=x^{3}+ax^{2}+bx+10\), where \(a\) and \(b\) are constants. The graph of \(y=f(x)\) has a point of inflexion at \(x=2\) and a stationary point at \(x=-1\). [13]

1. (a) (a) Find the value of \(a\) and the value of \(b\). [4]

   • M1 for \(f''(x)=6x+2a\)
   • A1 for \(6(2)+2a=0\), so \(a=-6\)
   • M1 for \(f'(x)=3x^{2}+2ax+b\) and \(f'(-1)=0\)
   • A1 for \(b=-15\)

2. (b) (b) Find the coordinates of the stationary points of the graph of \(y=f(x)\). Use the second derivative to decide the nature of each. [5]

   • M1 for \(f'(x)=3x^{2} - 12x - 15=0\) solved
   • A1 for \(x=-1\) and \(x=5\)
   • A1 for \((-1,18)\) and \((5,-90)\)
   • M1 for \(f''(x)=6 \left(x - 2\right)\) evaluated at both points
   • A1 for \(f''(-1)=-18<0\) (maximum) and \(f''(5)=18>0\) (minimum)

3. (c) (c) Write down the set of values of \(x\) for which the graph is concave down. Explain why there is a point of inflexion at \(x=2\). [2]

   • A1 for \(x<2\) (\(f''(x)<0\))
   • R1 for \(f''\) changes sign at \(x=2\)

4. (d) (d) Find the set of values of \(k\) for which \(f(x)=k\) has exactly three solutions. [2]

   • A1 for the values lie between the local minimum and the local maximum: \(-90<k<18\)
   • A1 for strict inequalities (at \(k=18\) or \(k=-90\) there are only two solutions)

A paint company makes closed cylindrical tins of volume 1000 cm³. A tin has base radius \(r\) cm and height \(h\) cm. The company wants to use as little metal as possible. [12]

1. (a) (a) Write down an expression for \(h\) in terms of \(r\). [1]

   • A1 for \(V=\pi r^{2}h=1000\) rearranged: \(h=\frac{1000}{\pi r^{2}}\)

2. (b) (b) Show that the total surface area is \(A=2\pi r^{2}+\frac{2000}{r}\). [2]

   • M1 for \(A=2\pi r^{2}+2\pi rh\)
   • A1 for \(2\pi r\times\frac{1000}{\pi r^{2}}=\frac{2000}{r}\) (AG)

3. (c) (c) Find \(\frac{dA}{dr}\). [2]

   • M1 for rewriting \(\frac{2000}{r}=2000r^{-1}\)
   • A1 for \(4 \pi r - \frac{2000}{r^{2}}\)

4. (d) (d) Find the value of \(r\) that minimizes the surface area. Give your answer to 3 significant figures. [3]

   • M1 for \(4 \pi r - \frac{2000}{r^{2}}=0\)
   • A1 for \(r^{3}=\frac{500}{\pi}\)
   • A1 for \(r=5.42\)

5. (e) (e) Use the second derivative to justify that this value of \(r\) gives a minimum. [2]

   • M1 for \(\frac{d^{2}A}{dr^{2}}=4 \left(\pi + \frac{1000}{r^{3}}\right)\)
   • R1 for positive for all \(r>0\), so a minimum (value 37.7 at \(r=5.42\))

6. (f) (f) Find the minimum surface area. [2]

   • M1 for substituting \(r=5.42\) into \(A\)
   • A1 for \(554\)

The number of members, \(P\), of an online chess club \(t\) months after it opens is modelled by \(P(t)=\frac{4500}{1+14e^{-0.7t}}\), \(t\geq 0\). [13]

1. (a) (a) Find the initial membership at \(t=0\). [2]

   • M1 for substituting \(t=0\): \(\frac{4500}{1+14}\)
   • A1 for \(300\)

2. (b) (b) Write down the value that \(P\) approaches as \(t\to\infty\). [1]

   • A1 for \(4500\)

3. (c) (c) Find \(\frac{dP}{dt}\). [3]

   • M1 for writing \(4500\left(1+14e^{-0.7t}\right)^{-1}\) and using the chain rule
   • A1 for \(4500\times(-1)\left(1+14e^{-0.7t}\right)^{-2}\) (outer function)
   • A1 for \(\frac{dP}{dt}=\frac{44100e^{-0.7t}}{\left(1+14e^{-0.7t}\right)^{2}}\)

4. (d) (d) Find the rate at which the membership is changing when \(t=3\). [2]

   • M1 for substituting \(t=3\) into \(\frac{dP}{dt}\)
   • A1 for \(733\)

5. (e) (e) Use your calculator to find the greatest rate of change of \(P\) and the value of \(t\) at which it occurs. [3]

   • M1 for a graph or table of \(\frac{dP}{dt}\) and finding its maximum
   • A1 for \(t=3.77\)
   • A1 for greatest rate \(788\)

6. (f) (f) Find the value of \(P\) at this time and describe what happens to the graph of \(P\) at this point. [2]

   • A1 for \(P=2250\) (half of the limiting value \(4500\))
   • R1 for a point of inflexion: the gradient is greatest here so \(P''=0\) and the concavity changes from up to down

A piston head moves in a straight line inside a cylinder. Its velocity at time \(t\) seconds is \(v\) m s\(^{-1}\), where \(v=4\sin(1.5t)\) for \(t\ge0\). The displacement of the piston head from a fixed point O is \(s\) metres, and \(s=2\) when \(t=0\). [12]

1. (a) (a) Find an expression for the acceleration. [2]

   • M1 for differentiating with the chain rule, factor \(1.5\)
   • A1 for \(a=6\cos(1.5t)\)

2. (b) (b) Find the first time \(t>0\) when the piston head is momentarily at rest. [2]

   • M1 for \(v=0\), i.e. \(\sin(1.5t)=0\) (radian mode)
   • A1 for \(t=2.09\)

3. (c) (c) Write down the greatest speed of the piston head. [1]

   • A1 for \(4\)

4. (d) (d) Find an expression for \(s\) in terms of \(t\). [2]

   • M1 for integrating \(v\) with respect to \(t\)
   • A1 for \(s=\frac{14}{3}-\frac{8}{3}\cos(1.5t)\) using \(s(0)=2\)

5. (e) (e) Find the displacement of the piston head from its starting position when \(t=5\). [2]

   • M1 for \(s(5)-s(0)\) or \(\int_0^{5}v\,dt\)
   • A1 for \(1.74\)

6. (f) (f) Find the total distance travelled by the piston head in the first 5 seconds. [3]

   • M1 for \(\int_0^{5}|v|\,dt\)
   • (M1) for the use of the absolute value, so that backward motion counts as positive
   • A1 for \(12.4\)