Lecture Note
University
Massachusetts Institute of TechnologyCourse
Multivariable CalculusPages
3
Academic year
2022
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Chain Rule Problem Solution To begin, we must first recall that the total differential of z equals the partialderivative of z in the x direction multiplied by dx plus the partial derivative of z in they direction multiplied by dy. π) ππ§ = π§ π₯ ππ₯ + π§ π¦ ππ¦ Now, looking at our formula for z, we see that the partial derivative in the x directionis 2x, and the partial derivative in the y direction is 2y. ππ§ = 2π₯ ππ₯ + 2π¦ ππ¦ π£ Next, we will solve for this partial derivative with respect to x using the chain rule.The chain rule states that whenever we compute a derivative, we can break downthe calculation into smaller steps by working on one term at a time and keeping trackof the relationships between them. So let's create a dependency graph. The equationbelow shows how the variables in my model are interrelated. At the top, we have z. zis a function of x and y, but x is itself a function of both u and v. And y is also afunction of u and v; that is, z depends on x and y. And x and y each jointly depend onu and vβthat is, none of these variables can be considered in isolation from anyothers because they all interact with one another. So, it's a little bit complicated, the relationships here. Now, what the chain rule saysis that, if we take a partial derivative of z with respect to u, we have to go through ourdependency graph every way that we can get from z to u. We get a term in oursummation for each one of those. For example, z goes to x goes to u. This meansthat we have partial z, partial x, partial x and partial u. Alternatively, we can also go zgoes to y goes to u , which gives us partial z, partial y and partial y. π) βπ§ βπ’ = βπ§ βπ₯ βπ₯βπ’ + βπ§ βπ¦ βπ¦βπ’ Now we will compute some of the partial derivatives. Partial z partial x is 2x, whichwe computed earlier. Partial x partial u is 2u; this result makes sense because x wasdefined as u squared minus v squared, so if we take a partial derivative in the udirection, we get 2u by definition. Similarly, partial y partial u is v again because ywas defined as uv; therefore, we take a partial derivative in the u direction to get v.
βπ§ βπ’ = 2π₯ * 2π’ + 2π¦ π£ So in fact, this is 4ux plus 2vy: βπ§ βπ’ = 4π’π₯ + 2π£π¦ And that's our partial derivative. So, notice that x is a function of u and v. So, wecould substitute an expression for x using its formula for u and v, but that's not reallynecessary. What's interesting about these problems is how the differentials dependon one another. And we`re perfectly happy with an answer that has mixed variableslike this. Now let us examine the problem under a new light. We will use a different method ofdifferentiation to arrive at the solution. Use total differentials, because the chain rulecan be more intuitively understood when using them. We will compute dz in this caseby differentiating both sides of the equation with respect to z. ππ§ = 2π₯ ππ₯ + 2π¦ ππ¦ Now we want to use the fact that x is itself a function of u and v. So that's what weneed to do now. Now, d dx equals 2u du minus 2v dv. ππ₯ = 2π’ ππ’ β 2π£ ππ£ And dy, so remember, y was. uv so taking d of uv, we get v du plus u dv. ππ¦ = π£ ππ’ + π’ ππ£ We now have all the partial derivatives, and we can move on to solving for z. Weonly need to substitute our formulas for dx into the formula for dz and divide by 2. ππ§ = 2π₯(2π’ ππ’ β 2π£ ππ£) So that was this term. And now we have plus 2y. V du plus u dv. ππ§ = 2π₯(2π’ ππ’ β 2π£ ππ£) + 2π¦(π£ ππ’ + π’ ππ£) It's just substitution, so now we just expand everything out using-- just expandingthese out. So if we collect all the things involving du, we have 4xu plus 2yv and thiswhole quantity times du. And then if we collect terms in dv, we have 2yu. So that'scoming from here. And then we have a minus 4xv. ππ§ = 2π₯ 2π’ ππ’ β 2π£ ππ£ ( ) + 2π¦ π£ ππ’ + π’ ππ£ ( ) = 4π₯π’ + 2π£π¦ ( )ππ’ + 2π¦π’ β 4π₯π£ ( )ππ£ And now we know that the partial derivative, written as dz/du or dz/dv, is thiscoefficient. So if we write the total differential as (dz/du)+(dz/dv), we can see that itequals the partial derivative of z with respect to u plus the partial derivative of z withrespect to v. ππ§ = βπ§ βπ’ ππ’ + βπ§ βπ£ ππ£
Let us consider an expression for parcels of the type z partial u. If we let thiscoefficient equal 4xu plus 2, one of these terms is an x, and one is a v, then we canfind that partial z partial u is equal to 4xu plus 2vy. βπ§ βπ’ = 4π₯π’ + 2π£π¦ ( ) Now, just as a sanity check to make sure we've covered this topic well, let's go backto the middle of the board and see that we got the same thing. So 4xu plus 2vy iswhat we concluded for partial z partial u. And now let's go over the two differentmethods and compare them. If you are in a rush, use the dependency graph along with tracing paths from z to u.By multiplying all of the partial derivatives that correspond to each edge and addingthem up, we get an expression. If you have more time, then use the method of totaldifferentials. This requires some basic calculus and some algebraic manipulations.As you saw, this method involved computing a greater number of derivatives thatwere not used in the final answer. However, believe that this method is moreconceptually straightforward than the first.
Solving Chain Rule Problems with Multiple Variables
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