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Find three different surfaces that contain the curve$ r(t) = t^2 i + \ln tj + (\frac{1}{t}) k $.

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$y=\ln \sqrt{x} \quad z=\frac{1}{\sqrt{x}} \quad z=e^{-y}$

01:06

Wen Zheng

Calculus 3

Chapter 13

Vector Functions

Section 1

Vector Functions and Space Curves

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Lectures

03:04

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x. The input of a function is called the argument and the output is called the value. The set of all permitted inputs is called the domain of the function. Similarly, the set of all permissible outputs is called the codomain. The most common symbols used to represent functions in mathematics are f and g. The set of all possible values of a function is called the image of the function, while the set of all functions from a set "A" to a set "B" is called the set of "B"-valued functions or the function space "B"["A"].

08:32

In mathematics, vector calculus is an important part of differential geometry, together with differential topology and differential geometry. It is also a tool used in many parts of physics. It is a collection of techniques to describe and study the properties of vector fields. It is a broad and deep subject that involves many different mathematical techniques.

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Find three different surfa…

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Identify the surface with …

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Verify that the curve $\ma…

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for this problem. We have a vector equation and we can write X equals t squared y equals the natural log of T and Z equals one virtue. So with that, we can write that y is equal to the natural log of T and we know that the natural log of T is just the natural log of the square root of t squared. That's just some way that we could write it, because now what we can do is say that that's the same thing as the square root of X. Then for Z, we can write that this is one over t but we know that that's the same thing as one over the square root of T squared which is just one over the square root of X. And then lastly, we can write Z as one over t which equals one over E to the y, which equals E to the negative one. So what we've seen is that by rewriting this, we can get things in terms of y and X. Uh, so this right here will be Our final equations are final solutions to the

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