Harder tangent and normal problems Problem set A One question at a time

Use technology to answer each part. Give each answer in the form \(y=mx+c\), with m and c to 3 significant figures.

1. (a) Find the equation of the tangent to \(y=x\sqrt{x+5}\) at \(x=4\).

   • From the calculator: derivative of \(x\sqrt{x+5}\) at \(x=4\) is \(3.667\)
   • \(y=12.00\) when \(x=4\)
   • \(y-12.00=3.667(x-4)\)
   • \(y=3.67x-2.67\)

2. (b) Find the equation of the normal to \(y=\frac{1}{x+6}\) at \(x=-4\).

   • From the calculator: derivative of \(\frac{1}{x+6}\) at \(x=-4\) is \(-0.2500\)
   • \(y=0.5000\) when \(x=-4\)
   • Normal gradient: \(m=-\frac{1}{-0.2500}=4.000\)
   • \(y-0.5000=4.000(x+4)\)
   • \(y=4.00x+16.5\)

Use technology to find the equation of the normal to each curve at the given point.

1. (a) \(y=x^2e^{3x}\) at \(x=0\)

   • From the calculator: the derivative at \(x=0\) is \(0\), so the tangent is horizontal
   • The point is \((0, 0)\). The normal is vertical: \(x=0\)

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

   • From the calculator: the derivative at \(x=2\) is \(0\), so the tangent is horizontal
   • The point is \((2, 0)\). The normal is vertical: \(x=2\)

Give each answer in the form \(y=mx+c\).

1. (a) Find the equation of the tangent to the curve \(y=x^2-3x\) which is parallel to the line \(y=5x+1\).

   • \(\frac{dy}{dx}=2x-3\). Parallel to the line, so the gradient is \(5\)
   • \(2x-3=5\) gives \(x=4\), and \(y=4\)
   • \(y-4=5(x-4)\)
   • \(y=5x-16\)

2. (b) Find the equation of the normal to the curve \(y=x^2-4x\) which is parallel to the line \(x+2y=6\).

   • \(\frac{dy}{dx}=2x-4\). The normal has gradient \(-\frac{1}{2}\), so the tangent has gradient \(2\)
   • \(2x-4=2\) gives \(x=3\), and \(y=-3\)
   • \(y+3=-\frac{1}{2}(x-3)\)
   • \(y=-\frac{1}{2}x-\frac{3}{2}\)

Find the equations of the normals to the curve \(y=x^3-3x\) which are perpendicular to the line \(y=9x+1\). Give each in the form \(y=mx+c\).

• The normal is perpendicular to the line, so the tangent is parallel to the line: \(\frac{dy}{dx}=9\)
• \(3x^{2}-3=9\) gives \(x=-2\) or \(x=2\)
• Points: \((-2, -2)\) and \((2, 2)\)
• Normal gradient \(-\frac{1}{9}\): \(y=-\frac{1}{9}x-\frac{20}{9}\) and \(y=-\frac{1}{9}x+\frac{20}{9}\)

Find the coordinates of the points on the curve \(y=x^3-3x^2+x\) at which the tangent is parallel to the line \(y=10x-4\).

• Parallel lines have equal gradients: \(\frac{dy}{dx}=10\)
• \(3x^{2}-6x+1=10\) gives \(x=-1\) or \(x=3\)
• Find y from the curve: \((-1, -5)\) and \((3, 3)\)

Find all the points on the curve \(y=x^2-2x-2\) where the gradient of the tangent equals the y-coordinate.

• \(\frac{dy}{dx}=2x-2\)
• Gradient = y: \(2x-2=x^2-2x-2\), so \(x^{2} - 4 x=0\)
• \(x=0\) or \(x=4\)
• Points: \((0, -2)\) and \((4, 6)\)

Find the coordinates of the point where the normal to each curve at the given point meets the curve again.

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

   • \(\frac{dy}{dx}=2x+1\), so at \(x=1\) the tangent gradient is \(3\) and \(y=2\)
   • Normal gradient \(-\frac{1}{3}\): \(y-2=-\frac{1}{3}(x-1)\), so \(y=-\frac{1}{3}x+\frac{7}{3}\)
   • Curve = normal: \(x^2+x=-\frac{1}{3}x+\frac{7}{3}\)
   • Multiply through and collect: \(3 x^{2} + 4 x - 7=0\)
   • \(x=1\) is a root (the point you started from). The other root is \(x=-\frac{7}{3}\)
   • \(y=\frac{28}{9}\), so the normal meets the curve again at \((-\frac{7}{3}, \frac{28}{9})\)

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

   • \(\frac{dy}{dx}=-2x^{-2}\), so at \(x=1\) the tangent gradient is \(-2\) and \(y=2\)
   • Normal gradient \(\frac{1}{2}\): \(y-2=\frac{1}{2}(x-1)\), so \(y=\frac{1}{2}x+\frac{3}{2}\)
   • Curve = normal: \(\frac{2}{x}=\frac{1}{2}x+\frac{3}{2}\)
   • Multiply through and collect: \(x^{2} + 3 x - 4=0\)
   • \(x=1\) is a root (the point you started from). The other root is \(x=-4\)
   • \(y=-\frac{1}{2}\), so the normal meets the curve again at \((-4, -\frac{1}{2})\)

A lamp is at the point \((3, -5)\). A ray of light leaves the lamp in a straight line that just touches the curve \(y=x^2+2\) (the line is a tangent to the curve).

1. (a) Find the coordinates of the two points where the line can touch the curve.

   • Let the point be \((p, q)\). Then \(q=p^{2}+2\)
   • \(\frac{dy}{dx}=2x\), so the gradient at \(x=p\) is \(2p\)
   • Tangent: \(y-q=(2p)(x-p)\)
   • It passes through \((3, -5)\): \(-5-(p^{2}+2)=(2p)(3-p)\)
   • Simplify (multiply through if there is a fraction): \(p^{2} - 6 p - 7=0\)
   • \(p=-1 \text{ or } p=7\)
   • Points: \((-1, 3)\) and \((7, 51)\)

2. (b) Write down the equation of the tangent at the point with the larger x-coordinate, in the form \(y=mx+c\).

   • Gradient at \(x=7\): \(2x=14\)
   • \(y-51=14(x-7)\)
   • \(y=14x-47\)

A curved cable is modelled by the curve \(y=\frac{a}{x}\), where a is a positive constant. The normal to the curve at the point where \(x=2\) is parallel to the line \(x-2y+3=0\). Find the value of a.

• \(y=ax^{-1}\), so \(\frac{dy}{dx}=-ax^{-2}\). At \(x=2\) the tangent gradient is \(-\frac{a}{4}\)
• Normal gradient: \(\frac{4}{a}\)
• The line \(x-2y+3=0\) has gradient \(\frac{1}{2}\)
• \(\frac{4}{a}=\frac{1}{2}\) gives \(a=8\)

The tangent to the curve \(y=\frac{6}{x}\) at the point where \(x=a\) (\(a>0\)) meets the y-axis at P and the x-axis at Q. O is the origin. Show that the area of triangle OPQ does not depend on a.

• At \(x=a\): \(y=\frac{6}{a}\) and gradient \(-\frac{6}{a^2}\)
• Tangent: \(y-\frac{6}{a}=-\frac{6}{a^2}(x-a)\)
• P is where \(x=0\): \(y=\frac{6}{a}+\frac{6}{a}=\frac{12}{a}\), so \(P=\left(0, \frac{12}{a}\right)\)
• Q is where \(y=0\): \(x-a=a\), so \(Q=(2a, 0)\)
• Area \(=\frac{1}{2}\times 2a\times\frac{12}{a}=12\)
• The area is the same for every a.

Show that the tangent to the curve \(y=x^3-5x+1\) at the point with x-coordinate k meets the curve again at the point with x-coordinate \(-2k\).

• \(\frac{dy}{dx}=3x^2-5\), so at \(x=k\) the gradient is \(3k^2-5\)
• At \(x=k\), \(y=k^3-5k+1\). Tangent: \(y=(3k^2-5)(x-k)+k^3-5k+1\)
• Curve = tangent: \(x^3-5x+1=(3k^2-5)(x-k)+k^3-5k+1\)
• \(x^3-3k^2x+2k^3=0\) (the other terms cancel)
• \((x-k)^2(x+2k)=0\), because expanding gives \(x^3-3k^2x+2k^3\)
• \(x=k\) is the point of contact (a repeated root). The other root is \(x=-2k\), so the tangent meets the curve again there.