Question

Find the slopes of the tangents to the curves of intersections of planes x=0,y=2 and the surface `z=x^3+e^(yx)` at point (0,2,1)?

Answer 1

We have a surface given by:

`z = x^3 + e^(yx)`

We are given the point of intersection as `(0,2,1)` so the intersection point of the planes `x=0` and `y=2` is superfluous, if we were not given the coordinates we obtain by substitution of `x=0` and `y=2` into the equation of the curve:

`z = 0 + e^(0) = 1`

First we rearrange the equation of the surface into the form `f(x,y,z)=0`

`z = x^3 + e^(yx) => x^3 + e^(yx) - z = 0`

And so we define our surface function, `f`, by:

`f(x,y,z) = x^3 + e^(xy) - z`

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

`bb(grad) f(x,y,z) = (partial f)/(partial x) bb(ul hat(i)) + (partial f)/(partial y) bb(ul hat(j)) + (partial f)/(partial z) bb(ul hat(k))`

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`bb(grad) f = (3x^2+ye^(xy)) bb(ul hat(i)) + (xe^(xy)) bb(ul hat(j)) + (-1) bb(ul hat(k))`

So for the particular point `(0,2,1)` the normal vector to the surface is given by:

`bb(grad) f(0,2,1) = (0+2e^0) bb(ul hat(i)) + (0) bb(ul hat(j)) + (-1) bb(ul hat(k))`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 bb(ul hat(i)) - bb(ul hat(k))`

So the tangent plane to the surface `z = x^3 + e^(yx)` at `(0,2,1)` has the normal vector `2 bb(ul hat(i)) - bb(ul hat(k))`

We can confirm this graphically: Here is the surface with the normal vector: