Harder tangent and normal problems Problem set B Worked solutions

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 \(y=x\sqrt{x+3}\) at \(x=5\).

   • From the calculator: derivative of \(x\sqrt{x+3}\) at \(x=5\) is \(3.712\)
   • \(y=14.14\) when \(x=5\)
   • \(y-14.14=3.712(x-5)\)
   • \(y=3.71x-4.42\)

2. (b) Find the equation of the normal to \(y=\frac{1}{x+7}\) at \(x=-5\).

   • From the calculator: derivative of \(\frac{1}{x+7}\) at \(x=-5\) is \(-0.2500\)
   • \(y=0.5000\) when \(x=-5\)
   • Normal gradient: \(m=-\frac{1}{-0.2500}=4.000\)
   • \(y-0.5000=4.000(x+5)\)
   • \(y=4.00x+20.5\)

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

1. (a) \(y=x^2e^{-4x}\) 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) \(y=(x-3)^2e^{x/2}\) at \(x=3\)

   • From the calculator: the derivative at \(x=3\) is \(0\), so the tangent is horizontal
   • The point is \((3, 0)\). The normal is vertical: \(x=3\)

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

1. (a) Find the equation of the tangent to the curve \(y=x^2-5x\) which is parallel to the line \(y=3x-7\).

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

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

   • \(\frac{dy}{dx}=2x-2\). The normal has gradient \(-\frac{1}{6}\), so the tangent has gradient \(6\)
   • \(2x-2=6\) gives \(x=4\), and \(y=8\)
   • \(y-8=-\frac{1}{6}(x-4)\)
   • \(y=-\frac{1}{6}x+\frac{26}{3}\)

Find the equations of the normals to the curve \(y=x^3-2x\) which are perpendicular to the line \(y=10x-3\). Give each in the form \(y=mx+c\).

• The normal is perpendicular to the line, so the tangent is parallel to the line: \(\frac{dy}{dx}=10\)
• \(3x^{2}-2=10\) gives \(x=-2\) or \(x=2\)
• Points: \((-2, -4)\) and \((2, 4)\)
• Normal gradient \(-\frac{1}{10}\): \(y=-\frac{1}{10}x-\frac{21}{5}\) and \(y=-\frac{1}{10}x+\frac{21}{5}\)

Find the coordinates of the points on the curve \(y=x^3-3x^2-2x\) at which the tangent is parallel to the line \(y=7x+5\).

• Parallel lines have equal gradients: \(\frac{dy}{dx}=7\)
• \(3x^{2}-6x-2=7\) gives \(x=-1\) or \(x=3\)
• Find y from the curve: \((-1, -2)\) and \((3, -6)\)

Find all the points on the curve \(y=x^2+4x-4\) where the gradient of the tangent equals the y-coordinate.

• \(\frac{dy}{dx}=2x+4\)
• Gradient = y: \(2x+4=x^2+4x-4\), so \(x^{2} + 2 x - 8=0\)
• \(x=-4\) or \(x=2\)
• Points: \((-4, -4)\) and \((2, 8)\)

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

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

   • \(\frac{dy}{dx}=2x+2\), so at \(x=1\) the tangent gradient is \(4\) and \(y=3\)
   • Normal gradient \(-\frac{1}{4}\): \(y-3=-\frac{1}{4}(x-1)\), so \(y=-\frac{1}{4}x+\frac{13}{4}\)
   • Curve = normal: \(x^2+2x=-\frac{1}{4}x+\frac{13}{4}\)
   • Multiply through and collect: \(4 x^{2} + 9 x - 13=0\)
   • \(x=1\) is a root (the point you started from). The other root is \(x=-\frac{13}{4}\)
   • \(y=\frac{65}{16}\), so the normal meets the curve again at \((-\frac{13}{4}, \frac{65}{16})\)

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

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

A rope is pulled tight from the point \((2, -2)\) so that it just touches the curved surface of a bowl with cross-section \(y=x^2+3\) (the rope lies along 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 \(q=p^{2}+3\)
   • \(\frac{dy}{dx}=2x\), so the gradient at \(x=p\) is \(2p\)
   • Tangent: \(y-q=(2p)(x-p)\)
   • It passes through \((2, -2)\): \(-2-(p^{2}+3)=(2p)(2-p)\)
   • Simplify (multiply through if there is a fraction): \(p^{2} - 4 p - 5=0\)
   • \(p=-1 \text{ or } p=5\)
   • Points: \((-1, 4)\) and \((5, 28)\)

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 \(x=5\): \(2x=10\)
   • \(y-28=10(x-5)\)
   • \(y=10x-22\)

A curved road is modelled by the curve \(y=\frac{a}{x}\), where a is a positive constant. The normal to the curve at the point where \(x=4\) is parallel to the line \(4x-y=9\). Find the value of a.

• \(y=ax^{-1}\), so \(\frac{dy}{dx}=-ax^{-2}\). At \(x=4\) the tangent gradient is \(-\frac{a}{16}\)
• Normal gradient: \(\frac{16}{a}\)
• The line \(4x-y=9\) has gradient \(4\)
• \(\frac{16}{a}=4\) gives \(a=4\)

The tangent to the curve \(y=\frac{10}{x}\) 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 \(x=a\): \(y=\frac{10}{a}\) and gradient \(-\frac{10}{a^2}\)
• Tangent: \(y-\frac{10}{a}=-\frac{10}{a^2}(x-a)\)
• P is where \(x=0\): \(y=\frac{10}{a}+\frac{10}{a}=\frac{20}{a}\), so \(P=\left(0, \frac{20}{a}\right)\)
• Q is where \(y=0\): \(x-a=a\), so \(Q=(2a, 0)\)
• Area \(=\frac{1}{2}\times 2a\times\frac{20}{a}=20\)
• The area is the same for every a.

Show that the tangent to the curve \(y=x^3+4x-7\) at the point with x-coordinate k meets the curve again at the point with x-coordinate \(-2k\).

• \(\frac{dy}{dx}=3x^2+4\), so at \(x=k\) the gradient is \(3k^2+4\)
• At \(x=k\), \(y=k^3+4k-7\). Tangent: \(y=(3k^2+4)(x-k)+k^3+4k-7\)
• Curve = tangent: \(x^3+4x-7=(3k^2+4)(x-k)+k^3+4k-7\)
• \(x^3-3k^2x+2k^3=0\) (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.