Lessons 10-18
Chain rule
First thing: create a tree diagram. Follow the branches to each potential path that leads to what you want ONLY, and then add each branch together. Also, don't forget to finish everything up by substituting any possible values.
Differentials
df=fx(dx)+fy(dy), which can also be expressed as fx(a,b)(x-a)+fy(a,b)(y-b). The latter is basically the exact same thing as the tangent plane/linear approximation equation EXCEPT you don't use the f(a,b).
Directional derivative
(gradient of f) dot (unit vector) This is basically the rate of change in a certain direction. Sometimes the vector that it will not give you is not unit vector! So make sure that you convert it before you finalize your answer.
Gradients and tangent planes
1. The function does not change at (a,b) in any direction orthogonal to the gradient of the function at the point. 2. The curve f(x,y)=k is a level curve, on which function values are constant. Through this, we can conclude that the gradient of the function at a point is orthogonal to the line TANGENT to the level curves.
Absolute extrema (Lagrange multiplers)
After you've found the critical points of some function, you can ONLY USE THIS METHOD FOR THE BOUNDARIES. Sometimes, it will give you a function that is a fixed constraint (aka, only a boundary line, you can spot this if the equation is something like x^2+y^2=1. Notice the use of =. This is what tells us we can only use Lagrange and move on. If it were an inequality, we'd have to use the critical points to find everything inside, and THEN use Lagrange on the outside). Just take the gradient of the given function, and set it to some arbitrary variable times the gradient of the constraints. Write everything out so it's a system of equations, with each partial equating to each other with the arbitrary variable. Solve the system of equations in order to get the extrema points.
Directional derivative, special case
Any unit vector can also be written as <costheta, sintheta>. In this instance, you would just use: fx(costheta)+fy(sintheta)
Gradient vector
Basically, you just form a vector based on the partial derivative for each corresponding point.
Finding the parametric equations for the line tangent to a 3D shape that intersects with a plane, given a point.
Find the gradients of each of them with the same mention as outlined above. Now, you want to take the cross product of the two gradients (at the given point), and the order of the cross product is just based on how the question outlines them. The result of the cross product will be the directional vector for the equation of the tangent plane, and you will use the given points as normal.
Finding the equation of a tangent plane and normal line to a surface given a point
First, you want to bump your given function up by a dimension. You can ensure your accuracy by just plugging in the given point, and making sure your bumped function turns out to 0. Now, bring it back down by taking the gradient of that function, and plug the point in at that gradient. Now, all you have to do is proceed as normal (like we did during Unit 1). Refer to the example 7b in the attached image.
Rate of change and directional derivatives
If a problem asks you to find the rate of change at a point in a direction, it's asking you to find the direction unit vector. If it asks for the direction, you just want the gradient. If it asks for the MOST RAPID INCREASE, you give the gradient. If it asks for the MOST RAPID DECREASE, you give -(gradient). If it wants the MAXIMUM RATE OF CHANGE, you should just take the magnitude of the gradient. if you want the MINIMUM RATE OF CHANGE, you do -|gradient|. In order to find the point of NO CHANGE, you want the gradient and unit vector to dot to 0, therefore you want them to be orthogonal. You can do this by just swapping positions on the gradient vector, and making one of them the opposite sign.
A rectangular box is resting on an xy plane with one vertex at the origin. The opposite vertex lies on some given plane. Find the maximum volume of such a box.
If problem shows up on the test, just skip it and come back to it in the end. It's very time consuming and annoying to solve this one. Basically, you take the volume equation, and similarly to above, substituent one of the variables with one from the given plane (you have to solve for it, usually). Now, you have an equation of two variables (just like above). Use the same method of finding the critical points by using the partials. Once you figure them out, you can plug them back into the volume formula to get the maximum volume.
Conditions for differentiability (differentiability implies continuity)
If the partial derivatives fx and fy exist near (a,b) and both are continuous at (a,b), then f is differentiable at (a,b). However, sometimes both partials at a point can exist, but f cannot be differentiable.
Finding the points on a surface that is closest to the origin.
In this situation, you use (x,y,z) as generic points, and (0,0,0) as your given point. Use the distance formula to find the distance, and square the whole thing to make your life easier. Now, you should have something like D=x^2+y^2+z^2. You want to maximize this, so you can usually use your given surface to plug in here and use that to simplify your life. Now, go ahead and find the critical points using the partial derivatives. Usually, it works out so that you can just use the SDT to determine what the minimum and maximum is.
Linear approximation
Literally the tangent plane formula, except now you're going to use that as a rough estimate to find the a certain value at a given point (x,y). Usually, (x,y) are the ugly numbers, while (a,b) are cleaner.
Second derivative test
Memorize this. If D>0, then you have a local max or min. If D<0, then it's a saddle point. If D=0, then the test is inconclusive. If fxx>0, it's a local minimum. If fxx<0 it's a local maximum.
Tangent plane equation
Memorize this. You can use the partial derivatives of a multivariable function in order to define it's tangent plane equation at a given POINT.
Multivariable implicit differentiation
Partial derivative of z in terms of x: -Fx/Fz Partial derivative of z in terms of y: -Fy/Fz
Finding relative maximum and minimum values
Remember that critical points are CANDIDATES for absolute extrema. You must never forget the endpoints. To find the critical points, just set both of the partial derivatives of a function simultaneously to 0. Use the system of equations to figure out which sets of points can make this system of equations true. This will give you the relative maxes and mins.
Finding the domain and range of multivariable functions
Simply base your response on the behaviors of the respective parent functions.
Partial derivatives
Simply hold one variable as a constant, while you derive the other. Sometimes you will have to take the partial derivative more than once. In this instance, simply refer to the exponent on the top of the differentiation symbol to denote the number of times something must be differentiated, and work from the closest to the f, outwards.
Absolute extrema (Calc I method)
So, we've figured out the critical points inside the function. But we also want to test the boundaries. You MUST use this method in certain circumstances, such as if you're given a square, a set of points that connect to a triangle, or a circle. After you find the critical points, take the boundary equation (the constraint), and solve it for one of the variables. Plug that into your main equation. Now you have an equation of one variable. Take this new equation, and solve it how you would in Calc I: find the critical points by taking the derivative at 0, and plug in the boundaries as well (boundaries within boundaries, basically). Now, compare everything to figure out your absolute max and min.
Limits and continuity, multivariable
The function is continuous at the point if the limit approaching the value is equal to the value at the point. This indicates that the function is defined at the point, the limit exists at the point, and the limit at the point is exactly at the value of the function.
Limit strategies
The same rules for the Calc I limits generally apply. There are five strategies you should always try: 1. Direct substitution 2. Convert to polar (r^2=x^2+y^2, x=rcostheta, y=rsintheta) 3. Multiply times the conjugate. 4. Factor. 5. Calculate the limit manually approaching some generic lines (x=0, y=0, x=y, etc) (You can also use Squeeze Theorem in instances such as finding the limit of cos(theta). Also, remember that the limit of sin(theta)/theta as the theta approaches 0 is 1.)
Lagrange multipliers with two constraints
This is basically the same thing, except now you have two constraints with two arbitrary variables in front of each of them, plus each other. If you have good algebra skills, this isn't very hard. If not, then good luck. gradient of F = (gradient of G)(variable 1) + (gradient of H)(variable 2)
Related rates
This is pretty much the exact same thing as in Calc I, HOWEVER this time you should expect to take the partial derivatives of the given formula. For example, if you get the volume of a cone, you have to find the partial of V. Use the tree, and don't forget to use your given values.
Finding the points on a given 3d shape that are closest to some point.
Use the distance formula, and minimize it to the point. The constant is the 3d shape, so it remains the same.
Finding the angle between two surfaces at a point.
Use the method outlined above to find the gradients of each of the surfaces, at the point. This will give you the normal vector to each surface. Now, just plug these two vectors into the formula from Unit 1.
Level curves
a line joining points of equal elevation on a surface. A collection of these is called a contour map.
Implicit Differentiation
dy/dx = -Fx/Fy Fy =/= 0
Rates of increase and decrease
f has its maximum rate of increase at a point. in the direction of the gradient at the point. The maximum rate of decrease at the same point is -(gradient at the point).
Clairaut's Theorem
fxy=fyx
