Tangents and normals by hand and with technology Problem set B 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) \(y=3x^2+2x-4\) at \(x=2\)

   • \(\frac{dy}{dx}=6x+2\)
   • When \(x=2\): \(m=14\) and \(y=12\)
   • \(y-12=14(x-2)\)
   • \(14x-y=16\)

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

   • \(\frac{dy}{dx}=3x^{2}-2x+5\)
   • When \(x=-2\): \(m=21\) and \(y=-22\)
   • \(y+22=21(x+2)\)
   • \(21x-y=-20\)

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

   • \(\frac{dy}{dx}=6x^{-2}\)
   • When \(x=3\): \(m=\frac{2}{3}\) and \(y=0\)
   • \(y=\frac{2}{3}(x-3)\)
   • \(2x-3y=6\)

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

1. (a) \(y=7-4x\) at \(x=1\)

   • \(\frac{dy}{dx}=-4\)
   • When \(x=1\): \(m=-4\) and \(y=3\)
   • Normal gradient: \(m=-1\div (-4)=\frac{1}{4}\)
   • \(y-3=\frac{1}{4}(x-1)\)
   • \(y=\frac{1}{4}x+\frac{11}{4}\)

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

   • \(\frac{dy}{dx}=3x^{2}+2\)
   • When \(x=1\): \(m=5\) and \(y=3\)
   • Normal gradient: \(m=-\frac{1}{5}\)
   • \(y-3=-\frac{1}{5}(x-1)\)
   • \(y=-\frac{1}{5}x+\frac{16}{5}\)

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

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

   • \(\frac{dy}{dx}=4x^{3}-6x\)
   • When \(x=-1\): \(m=2\) and \(y=-2\)
   • Normal gradient: \(m=-\frac{1}{2}\)
   • \(y+2=-\frac{1}{2}(x+1)\)
   • \(y=-\frac{1}{2}x-\frac{5}{2}\)

2. (b) \(y=3+\frac{4}{x}\) at \(x=2\)

   • \(\frac{dy}{dx}=-4x^{-2}\)
   • When \(x=2\): \(m=-1\) and \(y=5\)
   • Normal gradient: \(m=-1\div (-1)=1\)
   • \(y-5=(x-2)\)
   • \(y=x+3\)

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

   • \(\frac{dy}{dx}=3x^{2}-5\)
   • When \(x=2\): \(m=7\) and \(y=0\)
   • Normal gradient: \(m=-\frac{1}{7}\)
   • \(y=-\frac{1}{7}(x-2)\)
   • \(x+7y-2=0\)

2. (b) \(y=3x+\frac{4}{x}\) at \(x=2\)

   • \(\frac{dy}{dx}=3-4x^{-2}\)
   • When \(x=2\): \(m=2\) and \(y=8\)
   • Normal gradient: \(m=-\frac{1}{2}\)
   • \(y-8=-\frac{1}{2}(x-2)\)
   • \(x+2y-18=0\)

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) \(y=\sqrt{2x+1}\) at \(x=12\)

   • From the calculator: derivative of \(\sqrt{2x+1}\) at \(x=12\) is \(0.2000\)
   • \(y=5.000\) when \(x=12\)
   • \(y-5.000=0.2000(x-12)\)
   • \(y=0.200x+2.60\)

2. (b) \(y=6e^{-x}\) at \(x=2\)

   • From the calculator: derivative of \(6e^{-x}\) at \(x=2\) is \(-0.8120\)
   • \(y=0.8120\) when \(x=2\)
   • \(y-0.8120=-0.8120(x-2)\)
   • \(y=-0.812x+2.44\)

3. (c) \(y=\frac{1}{x^2+4}\) at \(x=1\)

   • From the calculator: derivative of \(\frac{1}{x^2+4}\) at \(x=1\) is \(-0.08000\)
   • \(y=0.2000\) when \(x=1\)
   • \(y-0.2000=-0.08000(x-1)\)
   • \(y=-0.0800x+0.280\)

The surface of a skate ramp has a cross-section modelled by \(y=x^3-9x^2+24x-10\), where x and y are in metres. A straight rail is fixed along the tangent to the ramp where \(x=3\), and a straight support 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.

   • \(\frac{dy}{dx}=3x^{2}-18x+24\)
   • When \(x=3\): \(m=-3\) and \(y=8\)
   • \(y-8=-3(x-3)\)
   • \(3x+y=17\)

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

   • \(\frac{dy}{dx}=3x^{2}-18x+24\)
   • When \(x=3\): \(m=-3\) and \(y=8\)
   • Normal gradient: \(m=-1\div (-3)=\frac{1}{3}\)
   • \(y-8=\frac{1}{3}(x-3)\)
   • \(y=\frac{1}{3}x+7\)

The concentration of a medicine in a patient's blood is modelled by \(C=6\ln(t+2)\), where C is in mg/L and t is the time in hours after the dose.

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

   • From the calculator: derivative of \(6\ln(t+2)\) at \(t=3\) is \(1.200\)
   • \(C=9.657\) when \(t=3\)
   • \(C-9.657=1.200(t-3)\)
   • \(C=1.20t+6.06\)

2. (b) Use the tangent to estimate C when \(t=4\).

   • Put \(t=4\) into the tangent: \(C=1.20\times 4+6.06\)
   • \(C\approx 10.9\) mg/L
   • The model itself gives 10.8 mg/L, so the tangent is a good estimate close to \(t=3\).

A normal is drawn to the curve \(y=2x^2+\frac{6}{x}-3\) 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.

   • \(\frac{dy}{dx}=4x-6x^{-2}\)
   • When \(x=1\): \(m=-2\) and \(y=5\)
   • Normal gradient: \(m=-1\div (-2)=\frac{1}{2}\)
   • \(y-5=\frac{1}{2}(x-1)\)
   • \(x-2y+9=0\)

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

   • The normal meets the x-axis where \(y=0\): \(x+9=0\)
   • \(x=-9\)

Use technology to answer these questions about the curve \(y=3x\ln x+1\).

1. (a) Find the equation of the tangent at \(x=2\), in the form \(y=mx+c\), with m and c to 3 significant figures.

   • From the calculator: derivative of \(3x\ln x+1\) at \(x=2\) is \(5.079\)
   • \(y=5.159\) when \(x=2\)
   • \(y-5.159=5.079(x-2)\)
   • \(y=5.08x-5.00\)

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

   • The tangent crosses the y-axis where \(x=0\).
   • \(y=c=−5.00\), so the point is \((0, −5.00)\)