If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

Question: 

`q001.  If f(x, y) = x^2 sqrt(y), and if x(t) = 2 t^3 while y(t) = t^2, t >= 0, then

• Find the expression for df/dt by using the chain rule df/dt = df/dx * dx/dt + df/dy * dy/dt (note that the 'd' symbols in df/dx and df/dy should be partial-derivative symbols.  The partial derivative symbol is a rounded-off 'd' that looks sort of like a backward '6').
• Evaluate your expression for t = 1.
• Substitute the expressions for x(t) and y(t) into the expression x^2 sqrt(y), to get f as a function of t only. 
• Find the derivative of this function.
• Evaluate your derivative at t = 1.

We first apply the chain rule df/dt = df/dx * dx/dt + df/dy * dy/dt:

df/dx = 2 x sqrt(y), df/dy = x^2 / (2 sqrt(y)), dx/dt = 6 t^2 and dy/dt = 2 t.  Thus

df/dt = df/dx * dx/dt + df/dy * dy/dt = 2 x sqrt(y) * 6 t^2 + x^2 / (2 sqrt(y)) * 2 t.

When t = 1 we easily see that x = 2 and y = 1, so our expression becomes

2 * 2 sqrt(1) * 6 * 1^2 + 2^2 / (2 sqrt(1)) * 2 * 1 = 24 + 4 = 28.

Substituting the expressions for x(t) and y(t) into the original function we get

f(x(t), y(t)) = (x(t))^2 sqrt(y(t)) = (2 t^3)^2 * sqrt(t^2) = 4 t^7. 

The derivative of this expression with respect to t is 28 t^6, which when t = 1 is 28.

The two results are the same, and this is no coincidence.  We can obtain the general expression for

Substituting the expressions for x(t) and y(t) into df/dt = 2 x sqrt(y) * 6 t^2 + x^2 / (2 sqrt(y)) * 2 t we get

df/dt = 2 ( 2 t^3) sqrt(t^2) * 6 t^2 + (2 t^3)^2 / (2 sqrt(t^2) * 2 t = 28 t^6.

This agrees with the expression 28 t^6, obtained previously by taking the derivative of f(x(t), y(t)) = 4 t^7.

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Question:

`q002.  If f(x, y) = e^x cos(y), and if x = u^2 - v^2 while y = u * v, then df = f_x dx + f_y dy = 2 x sqrt(y) dx + x^2 / (2 sqrt(y)) dy.

[ former statement was 'If f(x, y) = x^2 sqrt(y), and if x = u^2 - v^2 while y = u * v, then df = f_x dx + f_y dy = 2 x sqrt(y) dx + x^2 / (2 sqrt(y)) dy.' ]

• What is the expression for dx, in terms of u, v, du and dv?
• What is the expression for dy, in terms of u, v, du and dv?
• Substitute these expressions into df = 2 x sqrt(y) dx + x^2 / (2 sqrt(y)) dy to get an expression for df in terms of x, y, u, v, du and dv.
• Substitute the expressions for x and y directly into the resulting expression.

• Substitute the expressions for x and y in terms of u and v directly into the expression x^2 sqrt(y), obtaining an expression in just the variables u and v.
• Find the differential of your expression.
• Show that your differential is the same as the expression you obtained previously for df.

Your solution: 

Confidence rating:
 

Given Solution: 

x = u^2 - v^2 so dx = 2 u du - 2 v dv.

y = u * v so dy = du * v + v * du = v du + u dv

Substituting into df = e^x cos(y) dx - e^x sin(y) dy we get

df = e^(u^2 + v^2) cos( u v) ( 2 u du - 2 v dv) - e^(u^2 + v^2) sin( u v ) ( v du + u dv).

Substituting the expressions for x and y in terms of u and v directly into the expression f(x, y) = e^x cos(y) we obtain e^(u^2 - v^2) cos( u * v ). 

The derivative of this expression with respect to u is

df/du = 2 u e^(u^2 - v^2) cos(u v) - v e^(u^2 - v^2) sin ( u * v ). 

The derivative with respect to v is

df/dv = -2 v e^(u^2 - v^2) cos(u v) - u e^(u^2 - v^2) sin ( u * v ). 

The differential is therefore

df = (2 u e^(u^2 - v^2) cos(u v) - v e^(u^2 - v^2) sin ( u * v )) * du - ( 2 v e^(u^2 - v^2) cos(u v) + u e^(u^2 - v^2) sin ( u * v )) dv.

Our two expressions for df can be expanded and seen to agree term-for-term.

[ partial solution to former statement:

Substituting the expressions for x and y in terms of u and v we get

df = 2 (u^2 - v^2) sqrt(u v) ( 2 u du - 2 v dv ) + (u^2 - v^2)^2 / (2 sqrt( u v ) ) ( v du + u dv)

This isn't difficult to rearrange if we keep our bookkeeping straight:

df = 2 (u^2 - v^2) sqrt(u v) ( 2 u du - 2 v dv ) + (u^2 - v^2)^2 / (2 sqrt( u v ) ) ( v du + u dv)

df = (4 u (u^2 - v^2) sqrt(u v) + v (u^2 - v^2)^2 / (2 sqrt( u v ) ) ) du + (  (-4 v (u^2 - v^2) sqrt(u v) + u (u^2 - v^2)^2 / (2 sqrt( u v ) ) ) dv

 

 

 

Substituting the expressions for x and y in terms of u and v directly into the expression f(x, y) = x^2 sqrt(y) we obtain (u^2 - v^2)^2 sqrt( u * v ).  The derivative of this expression with respect to u is ( 2 u ( u^2 - v^2) sqrt( u * v) - v / (2 u sqrt( u * v) ) ) / (sqrt( u * v)^2

x = u^2 - v^2 so dx = 2 u du - 2 v dv.

y = u * v so dy = du * v + v * du = v du + u dv

Substituting into df = 2 x sqrt(y) dx + x^2 / (2 sqrt(y)) dy we get

df = 2 x sqrt(y) ( 2 u du - 2 v dv ) + x^2 / (2 sqrt(y) ) ( v du + u dv)

Substituting the expressions for x and y in terms of u and v we get

df = 2 (u^2 - v^2) sqrt(u v) ( 2 u du - 2 v dv ) + (u^2 - v^2)^2 / (2 sqrt( u v ) ) ( v du + u dv)

This isn't difficult to rearrange if we keep our bookkeeping straight:

df = 2 (u^2 - v^2) sqrt(u v) ( 2 u du - 2 v dv ) + (u^2 - v^2)^2 / (2 sqrt( u v ) ) ( v du + u dv)

df = (4 u (u^2 - v^2) sqrt(u v) + v (u^2 - v^2)^2 / (2 sqrt( u v ) ) ) du + (  (-4 v (u^2 - v^2) sqrt(u v) + u (u^2 - v^2)^2 / (2 sqrt( u v ) ) ) dv ]

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Question:

`q003.  Suppose that at a certain clock time t, the quantities x and y take values 2 and 5.  A change of .01 in the value of t causes x to change by .04 and y to change by .07. 

The temperature T in the immediate neighborhood of the (x, y) point (2, 5) on a thin sheet of metal changes by 60 degrees for every unit of displacement in the x direction, and by 20 degrees for every unit of displacement in the y direction. 

• By how many degrees does x change when t changes by .025?
• By how many degrees does y change when t changes by .025?
• By how many degrees does the temperature change due to the change in x?
• By how many degrees does the temperature change due to the change in y?
• By how many degrees does the temperature therefore change?

• What is the rate of change of x with respect to t?
• What is the rate of change of y with respect to t?
• What is the rate of change of temperature with respect to x?
• What is the rate of change of temperature with respect to y?
• What is the rate of change of temperature with respect to t?
• Write out the symbolic chain-rule expression for dT/dt, where T stands for temperature, in terms of T_x, T_y, dx/dt and dy/dt.  Explain how your expression symbolizes the process by which you answered the preceding questions.

x changes by .04 for units for every .01 unit change in t, so if t changes by .025, the value of x changes by .10.  This can be figured out by a simple proportionality, but it's important to see how this is an instance of rate of change:

If x changes by .04 for units for every .01 unit change in t, then the average rate of change of x with respect to t is `dx / `dt = .04 / .01 = 4.

This result `dx / `dt = 4 applies to the .01-unit t interval.  This approximation can also be applied with reasonably accuracy to other small intervals in the vicinity of this point.  So we can say that for an other `dt sufficiently close to .01, in the neighborhood of the original point, the change in x is

`dx = (average rate of change of x with respect to t) * `dt. 

which is approximately equal to

`dx = 4 * `dt.

For the present .025-unit t interval, our approximation is therefore

`dx = 4 * `dt = 4 * .025 = .10.

The same reasoning tells us that in this neighborhood `dy / `dt is close to .07 / .01 = 7, so that if t changes by .025, the value of y changes by approximate amount `dy = 7 * .025 =..175.

Now since x changes by about .10 and the temperature changes by about 60 degrees for every unit of x, we conclude that the temperature changes by about

`dT = 60 deg / unit * .10 unit = 6 deg.

Similarly since y changes by about .175 and the temperature changes by about 20 degrees for every unit of y, we conclude that the temperature changes by

`dT = 20 deg / unit * .175 unit = 3.5 deg.

Thus when t changes by .025, the values of x and y to change accordingly, causing a change in temperature.  The temperature changes by 6 deg due to the change in x, and by 3.5 degrees due to the change in y.  This results in a change of 6 deg + 3.5 deg = 9.5 deg in the temperature.

The relevant rates of change are as follows:

• The rate of change of x with respect to t is 4.
• The rate of change of y with respect to t is 7.
• The rate of change of  T with respect to x is 60.
• The rate of change of  T with respect to y is 20.

A change in t causes x to change at rate dx/dt = 4, and a change in x causes T to change at rate dT/dx = 60.  The rate of change of T with respect to t, due only to the change in the x coordinate, is therefore dT/dx * dx/dt = 4 * 60 = 240.

A change in t causes y to change at rate dy/dt = 7, and a change in x causes T to change at rate dT/dy = 20.  The rate of change of T with respect to t, due only to the change in the y coordinate, is therefore dT/dy * dx/dt = 7 * 20 = 140.

The net rate of change of T with respect to t is the sum of the rates due to the changes in the x and y coordinates.

dT/dt = dT/dx * dx/dt + dT/dy * dy/dt.

and

`dT = ( dT/dx * dx/dt + dT/dy * dy/dt. ) * `dt

For the present example, let's assume that t is measured in minutes, with x and y in units of centimeters.  Then dT/dx = 60 deg / cm, dT/dy = 20 deg / cm, dx/dt = 4 cm/min, dy/dt = 7 cm/min, and `dt = .025 min.  The numbers then work out more meanginfully as follows:

`dT = (60 deg / cm * 4 cm / min + 60 deg / cm * 7 cm / min) * .025 min.

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