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Chapter 10: Problem 1

Finding a Derivative In Exercises \(1-4,\) find \(d y / d x\) $$ x=t^{2}, y=7-6 t $$

Short Answer

Expert verified
The derivative dy/dx equals -3/t.

Step by step solution

01

Find dx/dt

Differentiate \(x = t^2\) with respect to t. This gives \(dx/dt = 2t\).
02

Find dy/dt

Next, differentiate \(y = 7 - 6t\) with respect to t. This gives \(dy/dt = -6\).
03

Compute dy/dx

Finally, use the formula \(dy/dx = (dy/dt) / (dx/dt)\) to find the value of dy/dx. Substituting the values of \(dy/dt\) and \(dx/dt\) from the previous steps, we get \(dy/dx = -6 / (2t)\), which simplifies to \(dy/dx = -3/t\).

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Key Concepts

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

Parametric Equations
Parametric equations define a set of definitions where the coordinates
  • \(x\)
  • \(y\)
depend on one or more variables, usually called parameters. In our example, both \(x\) and \(y\) are described as functions of \(t\), the parameter.

This means:
  • \(x = t^2\)
  • \(y = 7 - 6t\)
Using a parameter allows for describing curves in a plane without explicitly solving for \(y\) in terms of \(x\). For example, a circle can be represented as:
  • \(x = \cos(t)\)
  • \(y = \sin(t)\)
This is helpful for representing geometric shapes that are not functions in the standard \(y = f(x)\) sense.
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate of change of a quantity. It's the process used to find the derivative of a function. When you differentiate a function, you are essentially calculating its slope at any given point. In the context of our problem:
  • We first differentiate the parametric equation \(x = t^2\) to get \(dx/dt = 2t\).
  • Next, we differentiate \(y = 7 - 6t\) to obtain \(dy/dt = -6\).
Differentiation becomes especially interesting when dealing with parametric equations because it allows us to understand how both coordinates change with respect to a parameter, offering deeper insights into the curve鈥檚 behavior.
Chain Rule
The chain rule is a critical tool in calculus used to differentiate compositions of functions. When two functions are linked (or composed), the chain rule breaks this composition down to simplify the differentiation process.

For parametric equations, we use the chain rule to find \(dy/dx\) when both \(x\) and \(y\) depend on a parameter \(t\). The chain rule formula helps determine \(dy/dx\) by:
  • Calculating \(dy/dt\): the derivative of \(y\) with respect to \(t\).
  • Calculating \(dx/dt\): the derivative of \(x\) with respect to \(t\).
  • Using the formula \(dy/dx = (dy/dt) / (dx/dt)\) to combine these two rates of change.
This approach becomes necessary when dealing with curves that cannot be expressed as \(y = f(x)\). In our example, by using the chain rule, we found \(dy/dx = -3/t\), expressing the rate of change of \(y\) with respect to \(x\).

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Most popular questions from this chapter

Area of a Region In Exercises \(57-60\) , use the integration capabilities of a graphing utility to approximate, to two decimal places, the area of the region bounded by the graph of the polar equation. $$ r=\frac{3}{2-\cos \theta} $$

Finding the Area of a Polar Region In Exercises \(5-16\) , find the area of the region. Interior of \(r=1-\sin \theta\)

Finding the Area of a Polar Region Between Two Curves In Exercises \(35-42,\) use a graphing utility to graph \(h\) the polar equations. Find the area of the given region analytically. Common interior of \(r=3-2 \sin \theta\) and \(r=-3+2 \sin \theta\)

Writing In Exercises 33 and \(34,\) use a graphing utility to graph the polar equations and approximate the points of intersection of the graphs. Watch the graphs as they are traced in the viewing window. Explain why the pole is not a point of intersection obtained by solving the equations simultaneously. $$ \begin{array}{l}{r=4 \sin \theta} \\ {r=2(1+\sin \theta)}\end{array} $$

Finding the Area of a Polar Region In Exercises \(17-24\) , use a graphing utility to graph the polar equation. Find the area of the given region analytically. Between the loops of \(r=1+2 \cos \theta\)
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