Simple partial derivative question

1. Jun 11, 2013

leehufford

Hello, just a quick question about interpreting the partial derivative as a rate of change.

My example is the area of a parallelogram: A = absinθ, with a and b being the adjecent sides with θ being the angle between them.

We found the rate of change of the area A with respect to the side a by holding b and θ constant:

(for a = 10, b = 20, θ = ∏/6)

∂A/∂a = bsinθ

∂A/∂a = 20sin(∏/6) = 10.

Does this mean for every unit increment of a, A increases 10 times that, ie. increasing a by 2 increases A by 20?

Lee

2. Jun 11, 2013

SteamKing

Staff Emeritus
You can check this with the example in the OP. If a = 11, b = 20 and theta = pi/6, what is the area of your P-gram?

3. Jun 11, 2013

HallsofIvy

Here, your original function, $A= ab sin(\theta)$, is linear in both variables so, yes, you can think of the partial derivative as a "slope"- if a or b doubles, A also doubles.

If it were not linear, that would not be true. Take the example of a rectanguar solid with square cross sections- two perpendicular edges of length x and the third of length y. The volume is $V= x^2y$ so the two partial derivatives are $V_x= 2xy$ and $V_y= 2x^2$. If x= 2, y= 2 then V=(4)(2)= 8. If x= 2, y= 4 (so x remains the same while y doubles), V= (4)(4)= 16. Yes, the volume has doubled. But if x= 4, y= 2 (so y remains the same while x doubles), V= (16)(2)= 32. The volume has been multiplied by $2^2= 4$.

4. Jun 11, 2013

leehufford

Thanks for the replies.... can't believe I didn't think to just plug in some numbers and see it for myself.

Lee