0
$\begingroup$

Two lines are perpendicular if the product of their gradients is $-1$. I know this identity and have for a long time, but never understood why that was so.

Can I get an explanation a primary school student will understand?

$\endgroup$
3
  • 2
    $\begingroup$ How do you explain a first school student what a gradient is? $\endgroup$
    – N74
    Commented Nov 26, 2016 at 18:41
  • $\begingroup$ math.stackexchange.com/questions/519620/… math.stackexchange.com/questions/48713/… $\endgroup$
    – Bob
    Commented Nov 26, 2016 at 18:41
  • $\begingroup$ @n74 Give them arrows (made of paper, say), and have them place the arrows on a hillside according to the direction of steepest descent. (And that direction could be found by placing a tennis ball on the ground and seeing which way it rolls.) $\endgroup$
    – Théophile
    Commented Nov 27, 2016 at 1:25

4 Answers 4

1
$\begingroup$

The coordinate system has a right angle in it already: vertical lines (like the $y$ axis) are perpendicular to horizontal lines (like the $x$ axis). If we wish to have perpendicular lines, we may turn this pair of lines into position. Say we want one of the gradients to be $a/b$. We can take the $x$ axis and rotate it counterclockwise so it passes through $(b,a)$. Then we must also rotate the $y$ axis, and in doing so it will pass through $(-a,b)$. Our rotation has preserved the origin, so we can use the points straight up to find gradient, so the second line's gradient is $-b/a$. ... Which is equal to $-m^{-1}$. So the transformation that rotates from the axes to a pair of perpendicular lines maintains the product of gradients as $-1$.

$\endgroup$
0
$\begingroup$

Let the two lines have equations $y = f(x)$ and $y = g(x)$, and they cross at $x_0$, that is $f(x_0) = g(x_0) = y_0$.

We assume $f,g$ differentiable at $x_0$, so they both have tangent lines

$y = f'(x_0) x + f(x_0)$

$y = g'(x_0) x + g(x_0)$.

Letting $m_1 = f'(x_0)$, $m_2 = g'(x_0)$, $b_1 = f(x_0)$ and $b_2 = g(x_0)$ the tangent lines are

$y = m_1 x + b_1$

$y = m_2 x + b_2$

they can be parametrized by

$(x,y) = (x_0, y_0) + \lambda(1, m_1)$

$(x,y) = (x_0, y_0) + \lambda(1, m_2)$

So they have tangent vectors $v_1 = (1,m_1)$ and $v_2 = (1,m_2)$

These are perpendicular iff $v_1 \cdot v_2 = 0$, that is $(1,m_1) \cdot (1, m_2) = 0$, in other words $1 + m_1 m_2 = 0$, that is $m_1 m_2 = -1$, so the product of theyr gradients is $f'(x_0) g'(x_0) = -1$

$\endgroup$
0
$\begingroup$

Let $D_1$ and $D_2$ be the lines

$D_1\;:\; a_1x-y=b_1$

$D_2\;:\;a_2x-y=b_2$

they are perpendicular if the normal vectors are perpendicular. Which gives

$\vec{n_1}•\vec{n_2}=0$ with

$\vec{n_1}=(a_1,-1)$ and $\vec{n_2}=(a_2,-1)$.

$\implies a_1a_2+1=0$

but $a_1=\frac{dy}{dx}=$ gradient of $D_1$

and $a_2=\frac{dy}{dx}=$ gradient of$\;D_2$.

$\endgroup$
0
$\begingroup$

Explain that a hill that is very steep has a high gradient because it reaches very high, and a hill which ascends steadily has a low gradient because it reaches _low_er

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .