Curves and Surfaces in (ℝ^3): Partial Derivatives

Homework relating to partial derivatives. Solutions are provided.

Problem 1

Find the first and second partial derivatives of the function

f[x_,y_]=Log[x] + Cos[x y]

Barry's Solution

D[f[x,y],x]

D[f[x,y],y]

D[f[x,y],x,x]

D[f[x,y],x,y]

D[f[x,y],y,x]

D[f[x,y],y,y]

Problem 2

Find the first and second partial derivatives of the function

f[x_,y_]=Exp[ Cos[x y]]

Barry's Solution

D[f[x,y],x]

D[f[x,y],y]

D[f[x,y],x,x]

D[f[x,y],x,y]

D[f[x,y],y,x]

D[f[x,y],y,y]

Problem 3

Find the first and second partial derivatives of the function

f[x_,y_]=x/(x^2+y^2-1)

Barry's Solution

D[f[x,y],x]

D[f[x,y],y]

D[f[x,y],x,x]

D[f[x,y],x,y]

D[f[x,y],y,x]

D[f[x,y],y,y]

Problem 4

The temperature at a point (x,y) on a flat metal plate is given by T(x,y)=60/(1+(x^2)+(y^2)), where T is measured in degrees celsius and x and y in meters. Find the rate of change of the temperature  with respect to distance at the point (2,1) in the  x-direction and the y-direction. In which of these two directions is the temperature changing most rapidly? Is the temperature increasing or decreasing at this point? Note: This problem is setting the stage for directional derivatives, which are in Section 14.5.

Barry's Solution

The rate of change is given by the derivative. Therefore, we need the derivative evaluated at the point (2,1).

f[x_,y_]=60/(1+x^2+y^2)

D[f[x,y],x]

%/.{x->2,y->1}

D[f[x,y],y]

%/.{x->2,y->1}

So in the x-direction at the point (2,1) the temperature is decreasing by 20/3 degrees per meter, while in the y-direction the temperature is decreasing by 10/3 degrees per meter. The temperature is decreasing most rapidly  in the x-direction.

ContourPlot[f[x,y],{x,1,3},{y,0,2}]

Problem 5

The total resistance R produced by three conductors with resistances (Subscript[R, 1]), (Subscript[R, 2]), (Subscript[R, 3]) connnected in a parallel electrical circuit is given by the formula

(1/R)= (1/(Subscript[R, 1]))+ (1/(Subscript[R, 2]))+ (1/(Subscript[R, 3]))

Find ((∂ R)/(∂ (Subscript[R, 1]))).

Barry's Solution

You can do this by hand, or have Mathematica do it for you. I will use Mathematica to do it, first solving for R and then differentiating.

Solve[(1/R)== (1/R1)+ (1/R2)+ (1/R3),R]

D[R1*R2*R3/(R1*R2+R3*R2+R1*R3),R1]

Problem 6

In a study of frost penetration it was found that the temperature T at time t (measured in days) at a depth of x feet can be modelled by the function

T0+(e^(-k x)) T1 sin(((2 π t)/365)-k x)

where T0, T1, and k are positve constants.

(a) Find ((∂ T)/(∂ x)) and ((∂ T)/(∂ t))and explain their physical significance.
(b) Sketch the function when T0=0, T1=10 and k=0.2.
(c) When is the maximum rate of increase of the temperature over the course of a year at a depth of 7 feet? (Use the numerical values from part (b)).

Barry's Solution

T[x_,t_]=T0+T1* Exp[-k*x] Sin[2Pi/365*t-k*x]

D[T[x,t],x]

Physically, ((∂ T)/(∂ x))represents the rate of change of the temperature with respect to the distance below ground at a fixed moment in time.

D[T[x,t],t]

Physically, ((∂ T)/(∂ t))represents the rate of change of the temperature with respect to the time at a fixed distance below the ground.

Here is a sketch:

T[x_,t_]=0+10* Exp[-0.2*x] Sin[2Pi/365*t-0.2*x]

Plot3D[T[x,t],{x,0,6},{t,0,365}]

The further you are below ground, the less seasonal temperature variations above the surface effect the temperature in the ground.

At a depth of 7 feet, the function becomes:

T[7,t]

The rate of change is the derivative,

D[T[7,t],t]

And the time of the maximum rate of increase is found from setting the derivative of the rate of change equal to zero and solving for t:

Solve[D[T[7,t],{t,2}]==0,t]

Let's sketch to see if our number makes sense:

Plot[T[7,t],{t,0,365}]
Plot[T[7,t],{t,60,100}]

The maximum rate of increase occurs where the slope of the tangent line is greatest, which from our sketch appears to be around 80 days.