Gradient at a point Problem set B Worked solutions

Find \(\frac{dy}{dx}\) at the given value.

1. (a) \(y=2x^5\) when \(x=2\)

   • \(\frac{dy}{dx}=10x^{4}\)
   • \(\text{When }x=2:\ \frac{dy}{dx}=10(2)^{4}=160\)

2. (b) \(y=x^3-4x^2\) when \(x=5\)

   • \(\frac{dy}{dx}=3x^{2} - 8x\)
   • \(\text{When }x=5:\ \frac{dy}{dx}=3(5)^{2}-8(5)=35\)

Find the derivative at the given value.

1. (a) \(K=7-2n^{-1}\) when \(n=4\)

   • \(\frac{dK}{dn}=\frac{2}{n^{2}}\)
   • \(\text{When }n=4:\ \frac{dK}{dn}=\frac{2}{(4)^{2}}=0.125\)

2. (b) \(\omega=\phi^4-\phi^{-2}\) when \(\phi=0.5\)

   • \(\frac{d\omega}{d\phi}=4\phi^{3} + \frac{2}{\phi^{3}}\)
   • \(\text{When }\phi=0.5:\ \frac{d\omega}{d\phi}=4(0.5)^{3}+\frac{2}{(0.5)^{3}}=16.5\)

Find the gradient of the tangent to the curve.

1. (a) \(y=x^4-3x\) at the point where \(x=-2\)

   • \(\frac{dy}{dx}=4x^{3} - 3\)
   • \(\text{When }x=-2:\ \frac{dy}{dx}=4(-2)^{3}-3=-35\)

2. (b) \(y=5x^2-2x\) at the point where \(x=1.5\)

   • \(\frac{dy}{dx}=10x - 2\)
   • \(\text{When }x=1.5:\ \frac{dy}{dx}=10(1.5)-2=13\)

Find the gradient of the tangent to the curve.

1. (a) \(y=x^2+\frac{8}{x}\) at the point where \(x=4\)

   • \(\frac{dy}{dx}=2x - \frac{8}{x^{2}}\)
   • \(\text{When }x=4:\ \frac{dy}{dx}=2(4)-\frac{8}{(4)^{2}}=7.5\)

2. (b) \(y=2x^3-\frac{5}{x}\) at the point where \(x=0.5\)

   • \(\frac{dy}{dx}=6x^{2} + \frac{5}{x^{2}}\)
   • \(\text{When }x=0.5:\ \frac{dy}{dx}=6(0.5)^{2}+\frac{5}{(0.5)^{2}}=21.5\)

Find the rate of change by differentiating.

1. (a) How quickly does \(h=3r^4\) change as r changes, when \(r=1\)?

   • Rate of change = \(\frac{dh}{dr}\)
   • \(\frac{dh}{dr}=12r^{3}\)
   • \(\text{When }r=1:\ \frac{dh}{dr}=12(1)^{3}=12\)

2. (b) What is the rate of increase of Z with respect to k when \(k=-3\), if \(Z=-7k^2\)?

   • Rate of change = \(\frac{dZ}{dk}\)
   • \(\frac{dZ}{dk}=-14k\)
   • \(\text{When }k=-3:\ \frac{dZ}{dk}=-14(-3)=42\)

Find the values of x at which the curve has the given gradient.

1. (a) The curve \(y=x^4+3x\), gradient \(-29\).

   • \(\frac{dy}{dx}=4x^{3} + 3\)
   • \(4x^{3} + 3=-29\)
   • Solve for x: \(x=-2\)

2. (b) The curve \(y=5x+\frac{8}{x}\), gradient \(3\).

   • \(\frac{dy}{dx}=5 - \frac{8}{x^{2}}\)
   • \(5 - \frac{8}{x^{2}}=3\)
   • Solve for x: \(x=-2\) or \(x=2\)

Let \(g(x)=x^2+7x\).

1. (a) Find the average rate of change of g between x = 2 and x = 3.

   • \(g(2)=18,\ g(3)=30\)
   • \(\frac{30-18}{3-2}=12\)

2. (b) Find the average rate of change of g between x = 2 and x = 2.1, and between x = 2 and x = 2.01.

   • \(g(2.1)=19.11\), so \(\frac{19.11-18}{0.1}=11.1\)
   • \(g(2.01)=18.1101\), so \(\frac{18.1101-18}{0.01}=11.01\)

3. (c) Find \(g'(2)\). What do the answers to the two parts above approach?

   • \(g'(x)=2x+7\), so \(g'(2)=11\)
   • The averages 12, 11.1, 11.01 approach 11, the gradient at x = 2

The temperature \(\theta\) °C of the water in a kettle t minutes after the kettle is switched on is modelled by \(\theta=15+18t-t^2\), for \(0\leq t\leq8\).

1. (a) How quickly is the temperature changing when \(t=4\)? Give the units.

   • \(\frac{d\theta}{dt}=18-2t\)
   • \(t=4:\ \frac{d\theta}{dt}=18-2(4)=10\)
   • \(\theta\) is in °C and t in minutes, so the rate is in °C per minute.

2. (b) At what time is the temperature increasing at 4 °C per minute?

   • \(18-2t=4\)
   • \(2t=14\Rightarrow t=7\)

A boat uses F litres of fuel per hour when it travels at v km/h, where \(F=5v+\frac{72}{v}\), for \(v>0\). Find the speed at which F is increasing at a rate of 3 litres per hour for each extra km/h.

• \(F=5v+72v^{-1}\Rightarrow\frac{dF}{dv}=5-\frac{72}{v^2}\)
• \(5-\frac{72}{v^2}=3\Rightarrow v^2=36\)
• \(v=\pm6\)
• v must be positive, so \(v=6\) km/h.

The money M euros in a savings jar, d days after the jar was started, is modelled by \(M=0.5d^2+4d\) for the first 30 days.

1. (a) Find the average rate at which the jar fills between d = 10 and d = 12.

   • \(M(10)=90,\ M(12)=120\)
   • \(\frac{120-90}{12-10}=15\) euros per day

2. (b) Find the rate at which the jar is filling at d = 10.

   • \(\frac{dM}{dd}=d+4\)
   • At d = 10: \(10+4=14\) euros per day

3. (c) Explain why the two answers are different.

   • The rate is increasing with d
   • So the average over 10 to 12 is more than the rate at the start, 10

Find the values of x at which the tangent to the curve \(y=x^3-9x^2+2x\) has gradient 2.

• \(\frac{dy}{dx}=3x^{2} - 18x + 2\)
• \(3x^2-18x+2=2\Rightarrow3x^2-18x=0\)
• \(3x(x-6)=0\Rightarrow x=0\text{ or }x=6\)

The curves \(y=2x^3-x\) and \(y=9x^2-13x\) have the same gradient at two values of x. Find them.

• \(\frac{d}{dx}(2x^3-x)=6x^2-1\)
• \(\frac{d}{dx}(9x^2-13x)=18x-13\)
• \(6x^2-1=18x-13\Rightarrow x^2-3x+2=0\)
• \((x-1)(x-2)=0\Rightarrow x=1\text{ or }x=2\)