Trapezoid - Calculus of Several Variables

[doogmas]

First time posting on this forum so let's see how this goes!

Homework Statement

The base B of a trapezoid increases in length at the rate of 2cm/sec and the top T decreases in length at the rate of 1cm/sec. If the height h is increasing at the rate of 3cm/sec how rapidly is the area A changing when B = 30cm, T = 50cm and h=10cm.

Homework Equations

A=1/2*(B+T)*h

The Attempt at a Solution

I wasn't too sure how to start this.
I had the equation of ∇f(x,y,z) = 2x-y+3z (setting x=base, y=top, z=height). From lecture notes on a previous example it showed finding the dot product with P(a,b,c) so in this case P(30,50,10).
Following this through:
(2i-j+3k) ° ((x-50)i + (y-30)j + (z-10)k) = 0
=> 2x-y+3z-100=0 which leaves me with the equation I started with but with just -100. I don't think this is the right route as there is no where to go from here and I haven't used the area formula yet. Anyone with any nudges in the right direction?

 

[lippyka]

The correct solution is to first calculate the area with the given values: A = 1/2*(30+50)*10 = 350 cm^2. Then you can use the equation A=1/2*(B+T)*h to calculate the rate of change in the area: dA/dt = 1/2*(2dB/dt + dT/dt)*h + 1/2*(B+T)*dh/dt = (2 - 1)*10 + (30+50)*3 = 30 cm^2/sec