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Use the chain rule to find the indicated derivative(s). $$\begin{aligned}&g^{\prime}(t), \quad \text { where } \quad g(t)=f(x(t), y(t)), f(x, y)=\sqrt{x^{2}+y^{2}}\\\ &x(t)=\sin t, y(t)=t^{2}+2\end{aligned}$$

Short Answer

Expert verified
The derivative \(g'(t)\) is \(g'(t) = \frac{\sin t}{\sqrt{\sin^{2} t + (t^{2}+2)^{2}}} * \cos t + \frac{t^{2}+2}{\sqrt{\sin^{2} t + (t^{2}+2)^{2}}} * 2t\).

Step by step solution

01

Compute the partial derivatives of f(x, y)

First we have to compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). \(f(x,y) = \sqrt{x^{2} + y^{2}}\), thus, \(f_x(x,y) = \frac{x}{\sqrt{x^{2} + y^{2}}}\) and \(f_y(x,y) = \frac{y}{\sqrt{x^{2} + y^{2}}}\). These derivatives are computed using the chain rule for functions of more than one variable.
02

Calculate the derivatives of x(t) and y(t)

Next, calculate the derivatives of the functions \(x(t) = \sin t\) and \(y(t) = t^{2}+2\). Using basic derivatives rules, we get: \(x'(t) = \cos t\) and \(y'(t) = 2t\).
03

Apply the chain rule

Now that we have \(f_x\), \(f_y\), \(x'(t)\), and \(y'(t)\), we can use them to find \(g'(t)\). According to the chain rule for functions of more than one variable, \(g'(t) = f_x * x'(t) + f_y * y'(t)\). Substituting the known values, we get \(g'(t) = \frac{x}{\sqrt{x^{2} + y^{2}}} * \cos t + \frac{y}{\sqrt{x^{2} + y^{2}}} * 2t = \frac{\sin t}{\sqrt{\sin^{2} t + (t^{2}+2)^{2}}} * \cos t + \frac{t^{2}+2}{\sqrt{\sin^{2} t + (t^{2}+2)^{2}}} * 2t\) .

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In multivariable calculus, when we have functions with more than one variable, 'partial derivatives' play a crucial role. Imagine you're in a rolling field, and you want to measure how steeply the ground slopes in the north direction, disregarding any east-west slope. That's the essence of a partial derivative: it measures the rate at which a function changes as one variable moves, while the others remain fixed.

For the function \( f(x, y) = \sqrt{x^2 + y^2} \), we take partial derivatives with respect to \( x \) and \( y \). This tells us how the function value changes as we move along the x-axis or the y-axis exclusively. Namely, \( f_x(x, y) = \frac{x}{\sqrt{x^2 + y^2}} \) and \( f_y(x, y) = \frac{y}{\sqrt{x^2 + y^2}} \) are the slopes in the x and y directions respectively. It's important to grasp that partial derivatives are independent of each other, they measure change along an axis in a 'partial', isolated manner.
Multivariable Calculus
Branching out from the concept of partial derivatives, 'multivariable calculus' is the study of calculus applied to functions of several variables. It's like doing calculus in a space where you're not confined to just walking forward or backward, but can also move left, right, up, or down.

In multivariable calculus, we analyze surfaces and curves in three-dimensional space and beyond. For a two-variable function, such as \( f(x, y) = \sqrt{x^2 + y^2} \), this could represent a three-dimensional surface, and by using partial derivatives, we can determine how this surface slopes at any point. Techniques from this calculus are used in physics, engineering, and economics to model and solve problems involving multidimensional systems. The key is how changes in multiple variables can influence the overall outcome of functions describing natural phenomena, structures, or financial models.
Derivative of a Composition of Functions
The 'derivative of a composition of functions' becomes relevant when we deal with a function tucked inside another, like Russian nesting dolls. In calculus, this is where the chain rule comes into play. Specifically, in our example, we aren't just finding the derivative of \( f(x, y) \) alone; we're finding the derivative of \( g(t) = f(x(t), y(t)) \), where \( x \) and \( y \) themselves change with \( t \).

The chain rule tells us how to differentiate this layered situation. It's a formula that connects the derivatives of these intertwined functions. For the chain rule in multivariable calculus, you'll see it as \( g'(t) = f_x \times x'(t) + f_y \times y'(t) \). Each part of this sum corresponds to 'how much \( f \) changes with \( x \)' multiplied by 'how much \( x \) changes with \( t \)' plus 'how much \( f \) changes with \( y \)' multiplied by 'how much \( y \) changes with \( t \)'. It's a symphony of derivatives, with each instrument playing its part to give us the full rate of change of the composition at any moment in time.

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Most popular questions from this chapter

If your graphing utility can draw three-dimensional parametric graphs, compare the wireframe graph of \(z=\ln \left(x^{2}+y^{2}\right)\) with the parametric graph of \(x(r, t)=r \cos t, y(r, t)=r \sin t\) and \(z(r, t)=\ln \left(r^{2}\right)\)

Suppose that \(f(x, y)\) is a function with continuous secondorder partial derivatives. Consider the curve obtained by intersecting the surface \(z=f(x, y)\) with the plane \(y=y_{0}\) Explain how the slope of this curve at the point \(x=x_{0}\) relates to \(\frac{\partial f}{\partial x}\left(x_{0}, y_{0}\right) .\) Relate the concavity of this curve at the point \(x=x_{0}\) to \(\frac{\partial^{2} f}{\partial x^{2}}\left(x_{0}, y_{0}\right)\)

Use the result of exercise 44 to find an equation of the tangent plane to the parametric surface at the indicated point. \(S\) is the cylinder \(y^{2}=2 x\) with \(0 \leq z \leq 2 ;\) at \((2,2,1).\)

In example 4.6 of this chapter, we looked at a manufacturing process. Suppose that a gauge of 4 mm results from a gap of 4 mm, a speed of \(10 \mathrm{m} / \mathrm{s}\) and a temperature of \(900^{\circ} .\) Further, suppose that an increase in gap of 0.05 mm increases the gauge by \(0.04 \mathrm{mm},\) an increase in speed of \(0.2 \mathrm{m} / \mathrm{s}\) increases the gauge by \(0.06 \mathrm{mm}\) and an increase in temperature of \(10^{\circ}\) decreases the gauge by \(0.04 \mathrm{mm}\). Thinking of gauge as a function of gap, speed and temperature, find the direction of maximum increase of gauge.

Show that the vector \langle 100,-100\rangle is perpendicular to \(\nabla \sin (x+y) .\) Explain why the directional derivative of \(\sin (x+y)\) in the direction of \langle 100,-100\rangle must be zero. Sketch a wireframe graph of \(z=\sin (x+y)\) from the viewpoint \((100,-100,0) .\) Explain why you only see one trace. Find a viewpoint from which \(z=\sin (2 x-y)\) only shows one trace.

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