# Find parametric equations for the midpoint $P$ of the ladder

The following problem appears at MIT OCW Course 18.02 multivariable calculus.

The top extremity of a ladder of length $$L$$ rests against a vertical wall, while the bottom is being pulled away.

Find parametric equations for the midpoint $$P$$ of the ladder, using as a parameter the angle $$\theta$$ between the ladder and the ground (i.e., the $$x$$-axis).

And here is a sketch of the diagram for the problem.

We can find the parametric equations for the midpoint $$P$$ by finding the vector $$OP.$$ I write $$OP$$ as a sum of two vectors $$OP = OB + BP.$$

We know how to calculate the two vectors $$OB$$ and $$BP.$$ Using the Pythagorean Theorem, we can find the first component of $$OB$$ and the second one is obviously zero, hence $$OB = \langle L \cos \theta, 0 \rangle.$$

And by assuming that $$BP$$ is the radius of a circle with center at $$B,$$ we can find that $$BP = \left \langle \frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta \right \rangle.$$

So, we find that $$OP = \left \langle \frac{3L}{2} \cos \theta, \frac{L}{2} \sin \theta \right \rangle.$$

But the professor's solution is $$OP = \biggl \langle -\frac{L}{2} \cos \theta,\frac{L}{2} \sin \theta \biggr \rangle.$$

What's wrong with my solution?

When you write a vector $$\vec v = (v_1,v_2)$$ you're implicitly writing $$\vec v = v_1 \vec i + v_2 \vec j$$ for some linearly independent vectors $$\vec i ,\vec j.$$ You wrote

$$\vec {OB} = (L \cos \theta,0)$$

Depending on which vectors $$i,j$$ you are choosing this can be either true or false.

It seems to me quite natural to consider the line representing the wall as the $$y$$ axis and the $$x$$ axis as the line drawn by the floor along with the usual coordinate system.

You've correctly identified the distance between $$O$$ and $$B$$ as $$L \cos \theta$$. In the usual coordinates we then have $$\vec {OB} = (-L \cos \theta,0) \quad \text { and not } \quad (L \cos \theta ,0)$$ This gives you the desired result of $$\vec {OB} + \vec {BP} = (-L/2 \cos \theta , L/2 \sin \theta).$$

Regardless of all this, it seems to me that it is much easier to find the coordinates of $$B$$ and $$Q$$ (where the ladder touches the wall) and to take the average component wise than to determine $$\vec{BP}.$$

• Thank you, that's really a silly mistake, I know the easier solution but I wanted to solve it using this way but I made this mistake. – Kais Hasan Jun 3 at 17:56

By the Pythagorean Theorem, the length of the base of the triangle in the diagram is $$\langle L \cos \theta, 0 \rangle;$$ however, we must take into account that the vector $$OB$$ points in the negative $$x$$-direction, hence we have that $$OB = \langle -L \cos \theta, 0 \rangle.$$ Everything else you have written is correct, so we have that $$OP = OB + BP = \langle -L \cos \theta, 0 \rangle + \biggl \langle \frac L 2 \cos \theta, \frac L 2 \sin \theta \biggr \rangle = \biggl \langle -\frac L 2 \cos \theta, \frac L 2 \sin \theta \biggr \rangle.$$