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TimeRip496
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So where is the
Source: http://www.math.ubc.ca/~feldman/m226/taylor2d.pdf
Why is it not a function of t? I am still new to this so do you have something i can read on about this?blue_leaf77 said:##F(t) = f(x(t),y(t))##, so despite written to be a function of ##t##, the functional form of ##F(t)## is not written explicitly as a function of ##t##.
It's not an explicit function of ##t##. When you want to do chain rule, you have to pay attention on which variables are written explicitly, despite whether or not these variables are functions of yet another variable. For example take ##F(t) = xy^2## where ##x = \sqrt{t}## and ##y=t-2##. If you want to calculate ##dF/dt##, you can either first express ##x## and ##y## in terms of ##t## and then differentiate w.r.t. ##t## or let ##F## be expressed in ##x## and ##y## then use the chain ruleTimeRip496 said:Why is it not a function of t?
I guess this problem should belong to multivariate calculus.TimeRip496 said:do you have something i can read on about this?
Thanks a lot! I now understand.blue_leaf77 said:It's not an explicit function of ##t##. When you want to do chain rule, you have to pay attention on which variables are written explicitly, despite whether or not these variables are functions of yet another variable. For example take ##F(t) = xy^2## where ##x = \sqrt{t}## and ##y=t-2##. If you want to calculate ##dF/dt##, you can either first express ##x## and ##y## in terms of ##t## and then differentiate w.r.t. ##t## or let ##F## be expressed in ##x## and ##y## then use the chain rule
$$
\frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} .
$$
Both answers should be identical.
I guess this problem should belong to multivariate calculus.
In the linked pdf file, the author for some reason sets ##\Delta t = 1##.Svein said:I do not agree with the formula shown in the OP (dimensional analysis again). The picture states that [itex]\frac{d}{dt}x(t) = \Delta x [/itex] which is obviously wrong. If x is distance and t is time, it tries to assert that velocity equals a (short) distance. The correct statement is [itex]\frac{d}{dt}x(t) \cdot \Delta t= \Delta x [/itex].
I can agree with the second line, but the third line is pure nonsense.
The Taylor expansion of a function is a mathematical representation of the function in terms of its derivatives at a specific point. It is an infinite series that allows us to approximate the value of a function at a given point.
The chain rule is a fundamental rule in calculus that allows us to find the derivative of a composite function. The Taylor expansion involves taking derivatives of a function at a specific point, which is similar to the process involved in using the chain rule.
The chain rule is important in Taylor expansion because it allows us to calculate the derivatives of composite functions, which is a key step in finding the coefficients of the Taylor series. Without the chain rule, we would not be able to accurately represent a function using its derivatives.
The purpose of using Taylor expansion is to approximate the value of a function at a given point. This can be useful in situations where the function is difficult to evaluate directly or when we need a more accurate approximation than what is provided by other methods.
The error in Taylor expansion is calculated using the remainder term, which takes into account the difference between the actual value of the function and the value obtained using the Taylor series. This error term becomes smaller as we use more terms in the Taylor series, allowing for a more accurate approximation of the function.