i. Find dy/dx, which gives the slope of the tangent line to a point (x, y) on the curve.
At the point (3, 4), the slope of the tangent line is
and the slope of the normal line is
Both of these lines pass through the point (3, 4), so their equations are
ii. Find the x-intercepts of PQ and PR :
Then the vector starting at Q and terminating at P is equivalent to the vector
(though note that this vector starts at the origin)
and the vector starting at R and terminating at P is equivalent to
Now, suppose we treat these two vectors as vectors in 3D space. They both lie in the plane, so the z-coordinates are 0.
Recall the cross product identity,
where θ is the angle between vectors a and b. Here we know PQ and PR are perpendicular to one another so sin(θ) = sin(90°) = 1. The magnitude of their cross product corresponds to the area of the parallelogram spanned by them. Cut this area in half to get the area of ∆PQR.
The area is then