English, asked by haneetvincent3102, 7 months ago


1. Consider the parametric forms.
x=t+ and y=t-1/t of a curve
(a) Find the equation of the tangent at t=2
(b) Find the equation of the normal at t=2

Answers

Answered by mansibansal03
0

Explanation:

curve.

1.2.3. Use the equation for arc length of a parametric curve.

1.2.4. Apply the formula for surface area to a volume generated by a parametric curve.

Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve?

Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve (x(t),y(t)), then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.

Derivatives of Parametric Equations

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

x(t)=2t+3,y(t)=3t−4,−2≤t≤3.

The graph of this curve appears in Figure 1.16. It is a line segment starting at (−1,−10) and ending at (9,5).

A straight line from (−1, −10) to (9, 5). The point (−1, −10) is marked t = −2, the point (3, −4) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t – 4, and −2 ≤ t ≤ 3

Figure 1.16 Graph of the line segment described by the given parametric equations.

We can eliminate the parameter by first solving the equation x(t)=2t+3 for t:

x(t) = 2t+3 x−3 = 2t t =

x−3

2

.

Substituting this into y(t), we obtain

y(t) = 3t−4 y = 3(

x−3

2

)−4 y =

3x

2

9

2

−4 y =

3x

2

17

2

.

The slope of this line is given by

dy

dx

=

3

2

. Next we calculate x′(t) and y′(t). This gives x′(t)=2 and y′(t)=3. Notice that

dy

dx

=

dy/dt

dx/dt

=

3

2

. This is no coincidence, as outlined in the following theorem.

hope it helps

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