Tangents and normals by hand and with technology Problem set A Worked solutions

Find the equation of the tangent to each curve at the given point. Give your answer in the form \(ax+by=c\), where a, b and c are integers.

1. (a) \(y=2x^2-5x+1\) at \(x=3\)

   • \(\frac{dy}{dx}=4x-5\)
   • When \(x=3\): \(m=7\) and \(y=4\)
   • \(y-4=7(x-3)\)
   • \(7x-y=17\)

2. (b) \(y=x^3-2x^2+3x\) at \(x=-1\)

   • \(\frac{dy}{dx}=3x^{2}-4x+3\)
   • When \(x=-1\): \(m=10\) and \(y=-6\)
   • \(y+6=10(x+1)\)
   • \(10x-y=-4\)

3. (c) \(y=1+\frac{8}{x}\) at \(x=4\)

   • \(\frac{dy}{dx}=-8x^{-2}\)
   • When \(x=4\): \(m=-\frac{1}{2}\) and \(y=3\)
   • \(y-3=-\frac{1}{2}(x-4)\)
   • \(x+2y=10\)

Find the equation of the normal to each curve at the given point. Give your answer in the form \(y=mx+c\).

1. (a) \(y=5x-3\) at \(x=2\)

   • \(\frac{dy}{dx}=5\)
   • When \(x=2\): \(m=5\) and \(y=7\)
   • Normal gradient: \(m=-\frac{1}{5}\)
   • \(y-7=-\frac{1}{5}(x-2)\)
   • \(y=-\frac{1}{5}x+\frac{37}{5}\)

2. (b) \(y=x^3+x^2\) at \(x=1\)

   • \(\frac{dy}{dx}=3x^{2}+2x\)
   • When \(x=1\): \(m=5\) and \(y=2\)
   • Normal gradient: \(m=-\frac{1}{5}\)
   • \(y-2=-\frac{1}{5}(x-1)\)
   • \(y=-\frac{1}{5}x+\frac{11}{5}\)

Find the equation of the normal to each curve at the given point. Give your answer in the form \(y=mx+c\).

1. (a) \(y=x^4-x^2\) at \(x=1\)

   • \(\frac{dy}{dx}=4x^{3}-2x\)
   • When \(x=1\): \(m=2\) and \(y=0\)
   • Normal gradient: \(m=-\frac{1}{2}\)
   • \(y=-\frac{1}{2}(x-1)\)
   • \(y=-\frac{1}{2}x+\frac{1}{2}\)

2. (b) \(y=2-\frac{6}{x}\) at \(x=3\)

   • \(\frac{dy}{dx}=6x^{-2}\)
   • When \(x=3\): \(m=\frac{2}{3}\) and \(y=0\)
   • Normal gradient: \(m=-1\div \frac{2}{3}=-\frac{3}{2}\)
   • \(y=-\frac{3}{2}(x-3)\)
   • \(y=-\frac{3}{2}x+\frac{9}{2}\)

Find the equation of the normal to each curve at the given point. Give your answer in the form \(ax+by+c=0\), where a, b and c are integers.

1. (a) \(y=x^3-4x+1\) at \(x=2\)

   • \(\frac{dy}{dx}=3x^{2}-4\)
   • When \(x=2\): \(m=8\) and \(y=1\)
   • Normal gradient: \(m=-\frac{1}{8}\)
   • \(y-1=-\frac{1}{8}(x-2)\)
   • \(x+8y-10=0\)

2. (b) \(y=x+\frac{5}{x}\) at \(x=1\)

   • \(\frac{dy}{dx}=1-5x^{-2}\)
   • When \(x=1\): \(m=-4\) and \(y=6\)
   • Normal gradient: \(m=-1\div (-4)=\frac{1}{4}\)
   • \(y-6=\frac{1}{4}(x-1)\)
   • \(x-4y+23=0\)

Use technology to find the equation of the tangent to each curve at the given point. Give your answer in the form \(y=mx+c\), with m and c to 3 significant figures.

1. (a) \(y=\sqrt{3x+1}\) at \(x=8\)

   • From the calculator: derivative of \(\sqrt{3x+1}\) at \(x=8\) is \(0.3000\)
   • \(y=5.000\) when \(x=8\)
   • \(y-5.000=0.3000(x-8)\)
   • \(y=0.300x+2.60\)

2. (b) \(y=4e^{-x}\) at \(x=1\)

   • From the calculator: derivative of \(4e^{-x}\) at \(x=1\) is \(-1.472\)
   • \(y=1.472\) when \(x=1\)
   • \(y-1.472=-1.472(x-1)\)
   • \(y=-1.47x+2.94\)

3. (c) \(y=\frac{1}{x^2+1}\) at \(x=2\)

   • From the calculator: derivative of \(\frac{1}{x^2+1}\) at \(x=2\) is \(-0.1600\)
   • \(y=0.2000\) when \(x=2\)
   • \(y-0.2000=-0.1600(x-2)\)
   • \(y=-0.160x+0.520\)

A hillside has a cross-section modelled by \(y=x^3-6x^2+9x+2\), where x and y are in metres. A straight drainage pipe is laid along the tangent to the hillside where \(x=2\), and a straight post is fixed along the normal at that point.

1. (a) Find the equation of the tangent, in the form \(ax+by=c\), where a, b and c are integers.

   • \(\frac{dy}{dx}=3x^{2}-12x+9\)
   • When \(x=2\): \(m=-3\) and \(y=4\)
   • \(y-4=-3(x-2)\)
   • \(3x+y=10\)

2. (b) Find the equation of the normal, in the form \(y=mx+c\).

   • \(\frac{dy}{dx}=3x^{2}-12x+9\)
   • When \(x=2\): \(m=-3\) and \(y=4\)
   • Normal gradient: \(m=-1\div (-3)=\frac{1}{3}\)
   • \(y-4=\frac{1}{3}(x-2)\)
   • \(y=\frac{1}{3}x+\frac{10}{3}\)

The height of a seedling is modelled by \(h=9\ln(t+1)\), where h is in centimetres and t is the time in weeks after planting.

1. (a) Use technology to find the equation of the tangent at \(t=4\), in the form \(h=mt+c\), with m and c to 3 significant figures.

   • From the calculator: derivative of \(9\ln(t+1)\) at \(t=4\) is \(1.800\)
   • \(h=14.48\) when \(t=4\)
   • \(h-14.48=1.800(t-4)\)
   • \(h=1.80t+7.28\)

2. (b) Use the tangent to estimate h when \(t=5\).

   • Put \(t=5\) into the tangent: \(h=1.80\times 5+7.28\)
   • \(h\approx 16.3\) cm
   • The model itself gives 16.1 cm, so the tangent is a good estimate close to \(t=4\).

A normal is drawn to the curve \(y=3x^2+\frac{2}{x}-1\) at the point where \(x=1\).

1. (a) Find the equation of the normal, in the form \(ax+by+c=0\), where a, b and c are integers.

   • \(\frac{dy}{dx}=6x-2x^{-2}\)
   • When \(x=1\): \(m=4\) and \(y=4\)
   • Normal gradient: \(m=-\frac{1}{4}\)
   • \(y-4=-\frac{1}{4}(x-1)\)
   • \(x+4y-17=0\)

2. (b) Find the x-coordinate of the point where this normal crosses the x-axis.

   • The normal meets the x-axis where \(y=0\): \(x-17=0\)
   • \(x=17\)

Use technology to answer these questions about the curve \(y=2x\ln x+3\).

1. (a) Find the equation of the tangent at \(x=3\), in the form \(y=mx+c\), with m and c to 3 significant figures.

   • From the calculator: derivative of \(2x\ln x+3\) at \(x=3\) is \(4.197\)
   • \(y=9.592\) when \(x=3\)
   • \(y-9.592=4.197(x-3)\)
   • \(y=4.20x-3.00\)

2. (b) Write down the coordinates of the point where this tangent crosses the y-axis.

   • The tangent crosses the y-axis where \(x=0\).
   • \(y=c=−3.00\), so the point is \((0, −3.00)\)