Using implicit differentiation and the chain rule to relate changing quantities, including dtdy=dxdy⋅dtdx,
Given three variables x, y, and z,
Hence, given dzdx, we can find an expression for dzdy by calculating dxdy.
Implicit differentiation is when we differentiate both sides of an equation. It is helpful when we have an equation that cannot be simplified to y=f(x).
For example:
Differentiating both sides with respect to x and using the product rule:
By the chain rule, we know that (y2)′=2y⋅dxdy:
Now collecting dxdy terms:
So
Since dtdy=dxdy⋅dtdx, you may be asked to use implicit differentiation to find dxdy, then with a given dtdx and point, you can find dtdy.
Given the time rate of change of radius, length, height, or width of a three dimensional object, you may find the time rate of change of volume by taking the derivative of the volume equation.
Let L be the distance from the origin of a point with coordinates (x,y). Then, given dtdx and dtdy, we can find dtdL at a given point (x,y).
Using a given rate of change dtdx and trigonometry, we can calculate dxdθ, which can be used to find dtdθ.
Nice work completing Related Rates, here's a quick recap of what we covered:
Exercises checked off
Using implicit differentiation and the chain rule to relate changing quantities, including dtdy=dxdy⋅dtdx,
Given three variables x, y, and z,
Hence, given dzdx, we can find an expression for dzdy by calculating dxdy.
Implicit differentiation is when we differentiate both sides of an equation. It is helpful when we have an equation that cannot be simplified to y=f(x).
For example:
Differentiating both sides with respect to x and using the product rule:
By the chain rule, we know that (y2)′=2y⋅dxdy:
Now collecting dxdy terms:
So
Since dtdy=dxdy⋅dtdx, you may be asked to use implicit differentiation to find dxdy, then with a given dtdx and point, you can find dtdy.
Given the time rate of change of radius, length, height, or width of a three dimensional object, you may find the time rate of change of volume by taking the derivative of the volume equation.
Let L be the distance from the origin of a point with coordinates (x,y). Then, given dtdx and dtdy, we can find dtdL at a given point (x,y).
Using a given rate of change dtdx and trigonometry, we can calculate dxdθ, which can be used to find dtdθ.
Nice work completing Related Rates, here's a quick recap of what we covered:
Exercises checked off