Differentiation

Have a go first. Then open the working one line at a time.

1. A general point

   Let the point be \((p, q)\). Then \(q=p^{2}+5\)

   The point is unknown, so give it letters. It is on the curve, so q is found from p.

2. Tangent at that point

   Tangent: \(y-q=(2p)(x-p)\)

   The gradient is \(\frac{dy}{dx}\) with \(x=p\), which is \(2p\).

3. Use the given point

   It passes through \((1, -3)\): \(-3-(p^{2}+5)=(2p)(1-p)\)

   The tangent passes through \((1, -3)\): put \(x=1\) and \(y=-3\) into its equation.

4. Solve for p

   Simplify (multiply through if there is a fraction): \(p^{2} - 2 p - 8=0\); \(p=-2 \text{ or } p=4\)

   A quadratic in p. Solve by factorising or with the polynomial solver.

5. Points

   Points: \((-2, 9)\) and \((4, 21)\)

   Find q for each p using \(q=p^2+5\). Two tangents can be drawn from the point.

Have a go first. Then open the working one line at a time.

1. The normal

   \(\frac{dy}{dx}=2x-2\), so at \(x=3\) the tangent gradient is \(4\) and \(y=3\); Normal gradient \(-\frac{1}{4}\): \(y-3=-\frac{1}{4}(x-3)\), so \(y=-\frac{1}{4}x+\frac{15}{4}\)

   Find the tangent gradient and the point, then the normal as in the last lesson.

2. Curve equals normal

   Curve = normal: \(x^2-2x=-\frac{1}{4}x+\frac{15}{4}\)

   Where the normal meets the curve, both give the same y.

3. A quadratic

   Multiply through and collect: \(4 x^{2} - 7 x - 15=0\)

   Multiply by 4 to clear the fraction. Collect everything on one side.

4. The other root

   \(x=3\) is a root (the point you started from). The other root is \(x=-\frac{5}{4}\)

   You already know \(x=3\) is a solution, so the other root is the new point. Use the polynomial solver to check.

5. The point

   \(y=\frac{65}{16}\), so the normal meets the curve again at \((-\frac{5}{4}, \frac{65}{16})\)

   Put the new x into the curve.

Have a go first. Then open the working one line at a time.

1. Gradient of the line

   \(2x-y=7\) is \(y=2x-7\), so the gradient is 2

   Parallel lines have equal gradients.

2. Gradient equals 2

   \(2x-4=2\) gives \(x=3\), and \(y=0\)

   Set \(\frac{dy}{dx}=2\) and solve for x.

3. Equation

   \(y=2(x-3)\); \(y=2x-6\)

   Use the point of contact and the gradient 2.

Have a go first. Then open the working one line at a time.

1. Tangent gradient

   \(y=ax^{-1}\), so \(\frac{dy}{dx}=-ax^{-2}\). At \(x=3\) this is \(-\frac{a}{9}\)

   The constant a stays in the working.

2. Normal gradient

   \(m_{normal}=-1\div\left(-\frac{a}{9}\right)=\frac{9}{a}\)

   Turn the gradient upside down and change the sign.

3. Line gradient

   \(3x-y=4\) is \(y=3x-4\), gradient 3

   The normal is parallel to this line, so it has gradient 3.

4. Solve

   \(\frac{9}{a}=3\), so \(a=3\)

   A single equation in a.

Have a go first. Then open the working one line at a time.

1. Tangent at x = a

   At \(x=a\): \(y=\frac{4}{a}\) and gradient \(-\frac{4}{a^2}\); Tangent: \(y-\frac{4}{a}=-\frac{4}{a^2}(x-a)\)

   Keep a as a letter. Write \(y=4x^{-1}\), so \(\frac{dy}{dx}=-4x^{-2}\).

2. P and Q

   P is where \(x=0\): \(y=\frac{4}{a}+\frac{4}{a}=\frac{8}{a}\), so \(P=\left(0, \frac{8}{a}\right)\); Q is where \(y=0\): \(x-a=a\), so \(Q=(2a, 0)\)

   P is on the y-axis (\(x=0\)). Q is on the x-axis (\(y=0\)).

3. Area

   Area \(=\frac{1}{2}\times 2a\times\frac{8}{a}=8\)

   The a cancels, so the area is 8 whatever the value of a. Say this in a sentence.

Have a go first. Then open the working one line at a time.

1. Tangent at x = k

   \(\frac{dy}{dx}=3x^2-3\), so at \(x=k\) the gradient is \(3k^2-3\); At \(x=k\), \(y=k^3-3k+2\). Tangent: \(y=(3k^2-3)(x-k)+k^3-3k+2\)

   Keep k as a letter.

2. Curve equals tangent

   Curve = tangent: \(x^3-3x+2=(3k^2-3)(x-k)+k^3-3k+2\); \(x^3-3k^2x+2k^3=0\) (the other terms cancel)

   Subtract and collect. The terms in the gradient and the constant cancel to leave a short cubic.

3. Factorise

   \((x-k)^2(x+2k)=0\), because expanding gives \(x^3-3k^2x+2k^3\)

   The double factor \((x-k)^2\) is the tangent touching the curve at \(x=k\).

4. Conclusion

   \(x=k\) is the point of contact (a repeated root). The other root is \(x=-2k\), so the tangent meets the curve again there.

   End with a sentence. This is a check of a given result, not a proof of a general theorem.

Have a go first. Then open the working one line at a time.

1. Gradient

   From the calculator: derivative of \(3+x^2e^{-2x}\) at \(x=0\) is \(0\)

   The tangent is horizontal.

2. Why this matters

   \(-1\div 0\) is not a number

   A horizontal tangent has a vertical normal. You cannot use \(m_{normal}=-\frac{1}{m}\).

3. The point

   \(y=3+0=3\), so the point is \((0, 3)\)

   A vertical line through this point.

4. Answer

   \(x=0\)

   The normal is the line \(x=0\), the y-axis.