Can you determine which of these gradient functions (derivatives) corresponds to each pair of tangent and normal lines given above?
For each pair, can you determine which of the straight lines is the tangent and which is the normal?
Let’s take the first pair of lines, \(2y-x-3=0\) and \(y+2x+1=0\). We know that the gradients of the lines are \(\frac{1}{2}\) and \(-2\) respectively and that they intersect at the point \((-1,1)\).
At the point where the tangent and normal intersect, the curve must have the same gradient as the tangent line. So if we substitute \(x=-1\) into each gradient function we should find one with a value of either \(\frac{1}{2}\) or \(-2\).
\(2\times (-1)+5=3\)
\(15\times(-1)^2+12\times(-1)+3=0\)
\(\dfrac{-2}{(-1)^2}=-2\)
\(24(-1)^2-8\times(-1)+6=38\)
We can see that gradient function (3) matches, so the line \(y+2x+1=0\) is the tangent line and the other is the normal line at \((-1,1)\) to the curve. The gradient function is \(\dfrac{-2}{x^2}\).
| Normal: \(2y-x-3=0\) | Tangent: \(y+2x+1=0\) | Gradient function (3): \(\dfrac{-2}{x^2}\) |
We can follow the same ideas through for the remaining pairs of lines and gradient functions to get.
| Tangent: \(y-3x-2=0\) | Normal: \(3y+x-6=0\) | Gradient function (2): \(15x^2+12x+3\) |
| Tangent: \(y-8x+6=0\) | Normal: \(16y+2x+31=0\) | Gradient function (4): \(24x^2-8x+6\) |
| Normal: \(y+x+2=0\) | Tangent: \(y-x-2=0\) | Gradient function (1): \(2x+5\) |
