Tangents, normals and gradient with technology Problem set B One question at a time

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) \(y=x(\ln x)^2\) at \(x=3\)

   • From the calculator: derivative of \(x(\ln x)^2\) at \(x=3\) is \(3.404\)
   • \(y=3.621\) when \(x=3\)
   • \(y-3.621=3.404(x-3)\)
   • \(y=3.40x-6.59\)

2. (b) \(y=\frac{6}{1+x^2}\) at \(x=2\)

   • From the calculator: derivative of \(\frac{6}{1+x^2}\) at \(x=2\) is \(-0.9600\)
   • \(y=1.200\) when \(x=2\)
   • \(y-1.200=-0.9600(x-2)\)
   • \(y=-0.960x+3.12\)

3. (c) \(y=3^x\) at \(x=2\)

   • From the calculator: derivative of \(3^x\) at \(x=2\) is \(9.888\)
   • \(y=9.000\) when \(x=2\)
   • \(y-9.000=9.888(x-2)\)
   • \(y=9.89x-10.8\)

Use technology to find the equation of the normal 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) \(y=\sqrt{x+7}\) at \(x=2\)

   • From the calculator: derivative of \(\sqrt{x+7}\) at \(x=2\) is \(0.1667\)
   • \(y=3.000\) when \(x=2\)
   • Normal gradient: \(m=-\frac{1}{0.1667}=-6.000\)
   • \(y-3.000=-6.000(x-2)\)
   • \(y=-6.00x+15.0\)

2. (b) \(y=\frac{x^2+3}{\sqrt{x}}\) at \(x=9\)

   • From the calculator: derivative of \(\frac{x^2+3}{\sqrt{x}}\) at \(x=9\) is \(4.444\)
   • \(y=28.00\) when \(x=9\)
   • Normal gradient: \(m=-\frac{1}{4.444}=-0.2250\)
   • \(y-28.00=-0.2250(x-9)\)
   • \(y=-0.225x+30.0\)

Use technology to find the equation of the normal 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) \(y=\ln 4x\) at \(x=4\)

   • From the calculator: derivative of \(\ln 4x\) at \(x=4\) is \(0.2500\)
   • \(y=2.773\) when \(x=4\)
   • Normal gradient: \(m=-\frac{1}{0.2500}=-4.000\)
   • \(y-2.773=-4.000(x-4)\)
   • \(y=-4.00x+18.8\)

2. (b) \(y=(x+2)e^{x}\) at \(x=0\)

   • From the calculator: derivative of \((x+2)e^{x}\) at \(x=0\) is \(3.000\)
   • \(y=2.000\) when \(x=0\)
   • Normal gradient: \(m=-\frac{1}{3.000}=-0.3333\)
   • \(y-2.000=-0.3333x\)
   • \(y=-0.333x+2.00\)

Use technology to find the x-coordinate of the point on each curve that has the given gradient.

1. (a) \(y=\sqrt{x+2}\), gradient \(0.2\)

   • Graph y1 = d/dx(sqrt(x+2)) and y2 = 0.2
   • Intersection: x = 4.25

2. (b) \(y=e^{3x}\), gradient \(12\)

   • Graph y1 = d/dx(e^(3x)) and y2 = 12
   • Intersection: x = 0.462

Use technology to find the x-coordinate of every point on each curve that has the given gradient.

1. (a) \(y=\frac{1}{x+2}\), gradient \(-0.5\)

   • Graph y1 = d/dx(1/(x+2)) and y2 = -0.5
   • Intersection: x = −3.41, x = −0.586

2. (b) \(y=x^2-2\ln x\), gradient \(3\) (for \(x>0\))

   • Graph y1 = d/dx(x^2-2ln(x)) and y2 = 3
   • Intersection: x = 2.00

A graph is drawn of \(y=x^4-5x\).

1. (a) Find an expression for \(\frac{dy}{dx}\).

   • \(\frac{dy}{dx}=4x^{3}-5\)

2. (b) Write down the gradient of the tangent at \(x=0\).

   • \(\frac{dy}{dx}\) at \(x=0\): \(m=-5\)

3. (c) Hence find the equation of the normal to the curve at \(x=0\).

   • Normal gradient: \(m=-1\div (-5)=\frac{1}{5}\)
   • \(y=\frac{1}{5}x\)

Let \(f(x)=x^5+\frac{2}{x}\).

1. (a) Find \(f'(x)\).

   • Write \(f(x)\) as \(x^{5}+2x^{-1}\)
   • \(f'(x)=5x^{4}-2x^{-2}\)

2. (b) Find the gradient of the tangent at \(x=1\).

   • \(f'(1)=3\)

3. (c) Hence show that the tangent at \(x=1\) passes through the origin.

   • \(f(1)=3\), so the tangent is \(y-3=3(x-1)\)
   • \(y=3x\)
   • When \(x=0\), \(y=0\), so the tangent passes through the origin.

The distance a cyclist has travelled, in kilometres, is modelled by \(d=t\ln(t+3)\), where t is the time in hours after the start.

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

   • From the calculator: derivative of \(t\ln(t+3)\) at \(t=1\) is \(1.636\)
   • \(d=1.386\) when \(t=1\)
   • \(d-1.386=1.636(t-1)\)
   • \(d=1.64t-0.250\)

2. (b) Find the value of t when the rate of change of d is 2 km per hour.

   • The rate of change is the gradient. Graph y1 = d/dx(x·ln(x+3)) and y2 = 2
   • Intersection: t = 1.97

Use technology to answer these questions about the curve \(y=x+\frac{6}{\sqrt{x}}\).

1. (a) Find the x-coordinate of the point where the gradient of the curve is 0.5.

   • Graph y1 = d/dx(x+6/sqrt(x)) and y2 = 0.5
   • Intersection: x = 3.30

2. (b) Find the equation of the normal to the curve at this point, in the form \(y=mx+c\), with m and c to 3 significant figures.

   • At this point \(y=6.604\) and the gradient is 0.5
   • Normal gradient: \(m=-\frac{1}{0.5}=-2.000\)
   • \(y-6.604=-2.000(x-3.302)\)
   • \(y=-2.00x+13.2\)