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Use Theorem 12.7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d w / d t, \text { where } w=x y \sin z, x=t^{2}, y=4 t^{3}, \text { and } z=t+1$$

Short Answer

Expert verified
ANSWER: The derivative of w with respect to t is $$\frac{dw}{dt} = 20t^4 \sin(t+1) + 4t^5 \cos(t+1)$$

Step by step solution

01

Find the derivatives of x, y, and z with respect to t

To find the derivatives of x, y, and z with respect to t, we can use the power rule and the constant rule. Recall that x = t^2, y= 4t^3, and z = t + 1. Using the power rule, we find the following derivatives: $$\frac{dx}{dt} = 2t$$ Using the power rule again for y: $$\frac{dy}{dt} = 12t^2$$ Since z is the sum of two functions, t and 1, we will find the derivatives of each function separately and then add them. The derivative of t with respect to t is 1, and the derivative of the constant 1 is 0. $$\frac{dz}{dt} = 1 + 0 = 1$$
02

Apply the Chain Rule to find the derivative of w with respect to t

To find the derivative of w with respect to t, we will apply the Chain Rule. Recall that w = x * y * sin(z). Using the product rule and the chain rule, we can find the derivative of w as follows: $$\frac{dw}{dt} = \frac{dx}{dt}y \sin z + x \frac{dy}{dt} \sin z + x y \cos z \frac{dz}{dt} $$ Now, substitute the derivatives we found in step 1 and the original expressions for x, y, and z: $$\frac{dw}{dt} = (2t)(4t^3) \sin (t+1) + (t^2) (12t^2) \sin (t+1) + (t^2)(4t^3) \cos (t+1) (1) $$
03

Simplify the expression

Now, let's simplify the expression for the derivative of w with respect to t: $$\frac{dw}{dt} = 8t^4 \sin(t+1) + 12t^4 \sin(t+1) + 4t^5 \cos(t+1) $$ Since the first two terms have a common factor of t^4 * sin(t+1), we can factor out this common term: $$\frac{dw}{dt} = t^4 \sin(t+1)(8 + 12) + 4t^5 \cos(t+1) $$ Simplify further: $$\frac{dw}{dt} = 20t^4 \sin(t+1) + 4t^5 \cos(t+1) $$ Thus, the derivative of w with respect to t is: $$\frac{dw}{dt} = 20t^4 \sin(t+1) + 4t^5 \cos(t+1)$$

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Key Concepts

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

Chain Rule
The Chain Rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is a function that is composed of two or more functions. To understand why the Chain Rule is important, picture a function like a machine with different sections. The input goes through these sections in sequence, each one performing an operation.

Here's how the Chain Rule works: If you have a function that's inside another function, such as \(f(g(x))\), then the derivative of this function is obtained by multiplying the derivative of the outer function by the derivative of the inner function. Mathematically, this is expressed as:
  • \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)
This might sound complex, but it helps simplify the process of finding derivatives for complicated expressions. Consider the exercise where we found the derivative of \(w\) with respect to \(t\), with \(w\) being a product of various functions. We used the Chain Rule to unpack the \(\sin(z)\) term since \(z\) was a function of \(t\). This allows us to differentiate step by step, making the process manageable and straightforward.
Product Rule
The Product Rule is essential for finding the derivative of products of two or more functions. If you find yourself with a function that is composed of two multiplied terms, like \(u(x) \cdot v(x)\), you'll need the Product Rule.

To apply the Product Rule, you don't just differentiate each function separately and multiply them; instead, the derivative of the product is given by:
  • \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\)
Now, let's relate this back to our exercise. The expression \(w = x \, y \, \sin(z)\) was a product of three terms: \(x\), \(y\), and \(\sin(z)\). To find \(\frac{dw}{dt}\), we couldn't simply find the derivative of each part and multiply them. Instead, we used the Product Rule in conjunction with the Chain Rule to correctly differentiate each term in the product one at a time. This ensured that terms like \(x \, y\) were handled systematically without missing any parts of their contribution to the overall derivative.
Power Rule
The Power Rule is a straightforward formula used in calculus to find the derivatives of polynomials. It's particularly helpful when differentiating terms where the function is raised to a power.

The rule states that if you have a term in the form \(x^n\), its derivative is:
  • \(n \, x^{n-1}\)
This makes differentiating polynomial expressions quick and systematic, turning what could be a complex calculation into a simple one.
In our exercise, when we differentiated individual components like \(x = t^2\) and \(y = 4t^3\), we used the Power Rule. It allowed us to quickly determine \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 12t^2\), simplifying the wider problem. Understanding and applying the Power Rule effectively allows you to tackle polynomial derivatives with confidence.

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Most popular questions from this chapter

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. a. What is the equation of the plane \(P\) in which the curve lies? b. What is the angle between \(P\) and the \(x y\) -plane? c. Prove that the curve is an ellipse in \(P\).

Show that the plane \(a x+b y+c z=d\) and the line \(\mathbf{r}(t)=\mathbf{r}_{0}+\mathbf{v} t,\) not in the plane, have no points of intersection if and only if \(\mathbf{v} \cdot\langle a, b, c\rangle=0 .\) Give a geometric explanation of this result.

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

A clothing company makes a profit of \(\$ 10\) on its long-sleeved T-shirts and \(\$ 5\) on its short-sleeved T-shirts. Assuming there is a \(\$ 200\) setup cost, the profit on \(\mathrm{T}\) -shirt sales is \(z=10 x+5 y-200,\) where \(x\) is the number of long-sleeved T-shirts sold and \(y\) is the number of short-sleeved T-shirts sold. Assume \(x\) and \(y\) are nonnegative. a. Graph the plane that gives the profit using the window $$ [0,40] \times[0,40] \times[-400,400] $$ b. If \(x=20\) and \(y=10,\) is the profit positive or negative? c. Describe the values of \(x\) and \(y\) for which the company breaks even (for which the profit is zero). Mark this set on your graph.

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