12.9.9 Tangents, normals and gradient with technology · AI SL · 12.9: Differentiation

12.9.9

Tangents, normals and gradient with technology

12.9.9

Tangents, normals and gradient with technology

By the end of this section

Key points

When to use the calculator

Do not be put off by an unfamiliar function. For \(\sqrt{x}\), \(\ln x\), \(e^x\) and products such as \(xe^x\), the calculator gives the gradient at a point.
The equation of the tangent is still \(y-y_1=m(x-x_1)\). The calculator supplies m, and the curve supplies \(y_1\).
For the normal, use \(m_{normal}=-\frac{1}{m}\) with the same point.
Calculator values are decimals. Keep them in full until the last line, then give 3 significant figures unless told otherwise.

Where does the curve have gradient k?

The gradient is the derivative, so you want the x-values where \(f'(x)=k\).
Graph \(y_1=f'(x)\) using the calculator's derivative, graph the horizontal line \(y_2=k\), and find the intersections.
There may be more than one intersection. Give them all, unless the curve is only defined for some x, for example \(x>0\) for \(\ln x\).
Copy the sketch into your working: the derivative graph and the line, with the intersections marked.

By hand, then use the result

For sums of powers, differentiate by hand: write \(\frac{1}{x}\) as \(x^{-1}\), then lower each power by 1.
A question in several parts is a chain: derivative, then gradient at a point, then the line. Use each answer in the next part.
To show a result, state your working and finish with a sentence: the tangent has equation \(y=5x\), so it passes through \((0, 0)\).

On the calculator

Derivative at a point

Menu > Calc > Derivative at a point

Enter the function and the x-value.
The result is the gradient of the curve (and of the tangent) at that point.
Get the y-coordinate separately, by evaluating the function at the same x.

Screenshot and video to come

Gradient equal to a value: graph and intersect

In the graph screen enter y1 = d/dx(f(x), x) and y2 = k.
Choose an intersection point. Read the x-coordinate.
Look across the whole screen for further intersections, and use the window if the curve is cut off.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.

Tangent to a logarithm curve

Using technology, find the equation of the tangent to the curve \(y=3\ln x\) at the point where \(x=2\). Give your answer in the form \(y=mx+c\), with m and c to 3 significant figures.

Working

Gradient
From the calculator: derivative of \(3\ln x\) at \(x=2\) is \(1.500\)

Use the derivative at a point. Do not worry if you can already differentiate: the calculator is quicker and the answer is the same.
The point
\(y=3\ln 2=2.079\)

The tangent passes through \((2, 2.079)\).
Line
\(y-2.079=1.500(x-2)\)

Use \(y-y_1=m(x-x_1)\) with the unrounded readings.
Answer
\(y=1.50x-0.921\)

Round m and c to 3 significant figures.

Normal with a square root

Using technology, find the equation of the normal to the curve \(y=\frac{x^2+1}{\sqrt{x}}\) at the point where \(x=9\). Give your answer in the form \(y=mx+c\), with m and c to 3 significant figures.

Working

Tangent gradient
From the calculator: derivative of \(\frac{x^2+1}{\sqrt{x}}\) at \(x=9\) is \(4.481\)

The curve has a square root in the denominator. Let the calculator do the derivative.
The point
\(y=27.33\) when \(x=9\)

\(\frac{9^2+1}{\sqrt{9}}=\frac{82}{3}\).
Normal gradient
Normal gradient: \(m=-\frac{1}{4.481}=-0.2231\)

Perpendicular gradients multiply to \(-1\).
Answer
\(y=-0.223x+29.3\)

Write the line through the point with this gradient and make y the subject. Round at the end.

Where is the gradient 5?

Using technology, find the x-coordinates of the points on the curve \(y=\ln x+x^2\) where the gradient is 5.

Working

Graph the gradient
Graph y1 = d/dx(ln(x)+x^2) and y2 = 5

The gradient is the value of the derivative, so draw the derivative and the line for the gradient you want.
Intersect
Intersection: x = 0.219 and x = 2.28

Find every intersection on the screen. The curve is only defined for \(x>0\), and both answers satisfy this.
Answer
\(x=0.219\) or \(x=2.28\)

The graph shown is \(y=\frac{1}{x}+2x\), the derivative. Copy the sketch into your working.

Differentiate, then the normal

A graph is drawn of \(y=x^4-4x\). (a) Find \(\frac{dy}{dx}\). (b) Find the gradient of the tangent at \(x=0\). (c) Hence find the equation of the normal to the curve at \(x=0\).

Working

(a) Derivative
\(\frac{dy}{dx}=4x^{3}-4\)

Lower each power by 1.
(b) Gradient at x = 0
\(4(0)^3-4=-4\)

Substitute \(x=0\) into the derivative.
(c) Normal gradient
Normal gradient: \(m=-1\div (-4)=\frac{1}{4}\)

The point is \((0, 0)\), because \(y=0^4-4(0)=0\).
(c) Equation
\(y=\frac{1}{4}x\)

A line through the origin has no constant term.

Show the tangent goes through the origin

Let \(f(x)=2x^3+\frac{2}{x}\). (a) Find \(f'(x)\). (b) Find the gradient of the tangent at \(x=1\). (c) Hence show that the tangent at \(x=1\) passes through the origin.

Working

(a) Derivative
\(f(x)=2x^3+2x^{-1}\), so \(f'(x)=6x^{2}-2x^{-2}\)

Write \(\frac{2}{x}\) as \(2x^{-1}\) first.
(b) Gradient
\(f'(1)=6-2=4\)

This is the gradient of the tangent at \(x=1\).
(c) The tangent
\(f(1)=2+2=4\), so \(y-4=4(x-1)\), which gives \(y=4x\)

Find the point, then the line.
(c) Conclusion
When \(x=0\), \(y=0\). The tangent passes through \((0, 0)\).

State the conclusion. A line \(y=4x\) has no constant term, so it goes through the origin.

Menu

Differentiation

Tangents, normals and gradient with technology

By the end of this section

Key points

When to use the calculator

Where does the curve have gradient k?

By hand, then use the result

On the calculator

Derivative at a point

Gradient equal to a value: graph and intersect

Worked examples

Tangent to a logarithm curve

Working

Normal with a square root

Working

Where is the gradient 5?

Working

Differentiate, then the normal

Working

Show the tangent goes through the origin

Working

Choose a skill

Quiz