Tangents, normals and gradient with technology Problem set A Worked solutions

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

   • From the calculator: derivative of \(x^2\ln x\) at \(x=2\) is \(4.773\)
   • \(y=2.773\) when \(x=2\)
   • \(y-2.773=4.773(x-2)\)
   • \(y=4.77x-6.77\)

2. (b) \(y=\frac{10}{1+x^2}\) at \(x=3\)

   • From the calculator: derivative of \(\frac{10}{1+x^2}\) at \(x=3\) is \(-0.6000\)
   • \(y=1.000\) when \(x=3\)
   • \(y-1.000=-0.6000(x-3)\)
   • \(y=-0.600x+2.80\)

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

   • From the calculator: derivative of \(2^x\) at \(x=3\) is \(5.545\)
   • \(y=8.000\) when \(x=3\)
   • \(y-8.000=5.545(x-3)\)
   • \(y=5.55x-8.64\)

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) \(y=\sqrt{x+5}\) at \(x=4\)

   • From the calculator: derivative of \(\sqrt{x+5}\) at \(x=4\) is \(0.1667\)
   • \(y=3.000\) when \(x=4\)
   • Normal gradient: \(m=-\frac{1}{0.1667}=-6.000\)
   • \(y-3.000=-6.000(x-4)\)
   • \(y=-6.00x+27.0\)

2. (b) \(y=\frac{x^2+8}{\sqrt{x}}\) at \(x=4\)

   • From the calculator: derivative of \(\frac{x^2+8}{\sqrt{x}}\) at \(x=4\) is \(2.500\)
   • \(y=12.00\) when \(x=4\)
   • Normal gradient: \(m=-\frac{1}{2.500}=-0.4000\)
   • \(y-12.00=-0.4000(x-4)\)
   • \(y=-0.400x+13.6\)

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) \(y=\ln 5x\) at \(x=2\)

   • From the calculator: derivative of \(\ln 5x\) at \(x=2\) is \(0.5000\)
   • \(y=2.303\) when \(x=2\)
   • Normal gradient: \(m=-\frac{1}{0.5000}=-2.000\)
   • \(y-2.303=-2.000(x-2)\)
   • \(y=-2.00x+6.30\)

2. (b) \(y=(x+3)e^{x}\) at \(x=0\)

   • From the calculator: derivative of \((x+3)e^{x}\) at \(x=0\) is \(4.000\)
   • \(y=3.000\) when \(x=0\)
   • Normal gradient: \(m=-\frac{1}{4.000}=-0.2500\)
   • \(y-3.000=-0.2500x\)
   • \(y=-0.250x+3.00\)

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

1. (a) \(y=\sqrt{x+3}\), gradient \(0.1\)

   • Graph y1 = d/dx(sqrt(x+3)) and y2 = 0.1
   • Intersection: x = 22.0

2. (b) \(y=e^{2x}\), gradient \(10\)

   • Graph y1 = d/dx(e^(2x)) and y2 = 10
   • Intersection: x = 0.805

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

1. (a) \(y=\frac{1}{x+1}\), gradient \(-0.25\)

   • Graph y1 = d/dx(1/(x+1)) and y2 = -0.25
   • Intersection: x = −3.00, x = 1.00

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

   • Graph y1 = d/dx(x^2-ln(x)) and y2 = 3
   • Intersection: x = 1.78

A graph is drawn of \(y=x^4+2x\).

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

   • \(\frac{dy}{dx}=4x^{3}+2\)

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

   • \(\frac{dy}{dx}\) at \(x=0\): \(m=2\)

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

   • Normal gradient: \(m=-\frac{1}{2}\)
   • \(y=-\frac{1}{2}x\)

Let \(f(x)=2x^4+\frac{3}{x}\).

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

   • Write \(f(x)\) as \(2x^{4}+3x^{-1}\)
   • \(f'(x)=8x^{3}-3x^{-2}\)

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

   • \(f'(1)=5\)

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

   • \(f(1)=5\), so the tangent is \(y-5=5(x-1)\)
   • \(y=5x\)
   • When \(x=0\), \(y=0\), so the tangent passes through the origin.

The number of fish in a lake, in hundreds, is modelled by \(F=t\ln(t+1)\), where t is the time in years after the lake is stocked.

1. (a) Use technology to find the equation of the tangent at \(t=2\), in the form \(F=mt+c\), with m and c to 3 significant figures.

   • From the calculator: derivative of \(t\ln(t+1)\) at \(t=2\) is \(1.765\)
   • \(F=2.197\) when \(t=2\)
   • \(F-2.197=1.765(t-2)\)
   • \(F=1.77t-1.33\)

2. (b) Find the value of t when the rate of change of F is 1.5 hundred fish per year.

   • The rate of change is the gradient. Graph y1 = d/dx(x·ln(x+1)) and y2 = 1.5
   • Intersection: t = 1.47

Use technology to answer these questions about the curve \(y=x+\frac{4}{\sqrt{x}}\).

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

   • Graph y1 = d/dx(x+4/sqrt(x)) and y2 = 0.5
   • Intersection: x = 2.52

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 \(y=5.040\) and the gradient is 0.5
   • Normal gradient: \(m=-\frac{1}{0.5}=-2.000\)
   • \(y-5.040=-2.000(x-2.520)\)
   • \(y=-2.00x+10.1\)