1. 12.2 Differentiation and Integration of Vector-Valued FunctionsBy
Dr. Julia Arnold

2. In this section you will find that differentiation and integration of vector valued functions is term by term and subject to all the rules you have had previously.
Please look over the material before viewing this presentation.
Example 1: Sketch the plane curve represented by the vector-valued function, and sketch the vectors r(to) and r’(to) for the given value to. Position the vectors such that the initial point of r(to) is at the origin and initial point of r’(to) is at the terminal point of r(to) . What is the relationship between r’(to) and the curve?

3. r’(1) is tangent to curve at t = 2
Example 1: Sketch the plane curve represented by the vector-valued function, and sketch the vectors r(to) and r’(to) for the given value to. Position the vectors such that the initial point of r(to) is at the origin and initial point of r’(to) is at the terminal point of r(to) . What is the relationship between r’(to) and the curve?
Solution
r’(1) is tangent to curve at t = 2
Drawing done in WinPlot
View clip (if the clip won’t open in ppt. The file is included as graphpara.swf – open that
You will be asked if the video clip is a safe file. It is.)

4. Example 2: Sketch the space curve represented by the vector-valued function, and (b) sketch the vectors r(to) and r’(to) for the given value of to.
Solution:
The red curve is The Curve.
I picked points and then connected them.
The blue curve is a mirror image in the xy plane to give you perspective that the curve lies on the plane z = 3/2
The shorter looking vector is r(2) and the longer looking vector is r’(2) and is lying in the xy plane.
r(2) is longer than r’(2) in reality.
Drawing done in WinPlot Video Clip

5. Example 3: Find r’(t)
Solution:

6. Example 3: Find r’(t)
Solution:

7. Example 4: Find (a) r”(t) and (b) r’(t) . r”(t)
Solution:
(a)
(b)

8. Example 5: Find the open intervals on which the curve given by the vector-valued function is smooth.
Solution:
Since is continuous is continuous and is continuous for all real numbers so the first part is satisfied. Now we need to see if for some theta value.

9. Solution:
Here the sin function is 0 for
While 1 + cos is 0 for
So, both are 0 for
Thus r is smooth on
See graph on next slide

10. WinPlot drawing

11. Example 6: Use the properties of the derivative to find
for
Solution:
First we find the cross product. If you have never seen the short cut method for finding the determinant of a 3x3 matrix, then pay close attention to the method I use.
For explanation of my method of finding a determinant for a 3x3 matrix view clip.

12. Now we can find the derivative Dt
I used the quotient rule in the first line.
Factored out the 1/t2 in the second line before using the product rule.

13. Example 7: Find the indefinite integral
Solution:
Since we are involved with two major functions I would probably try parts and see if that would work:
This is just the answer for the i term.

14. Example 7: Find the indefinite integral
Solution:
Since we are involved with two major functions I would probably try parts and see if that would work:
This is just the answer for the j term.

15. =

16. Example 8: Find r(t) for the given conditions.
Solution:
This is called an initial value problem. We find the antiderivative realizing their could be arbitrary constants involved. Hopefully, the given condition will lead us to a solution which allows us to find those constants.
This concludes the presentation.

17. What is meant mathematically speaking by a smooth curve?
Answer:
The parameterization of the curve represented by the vector-valued function r(t)=f(t)i + g(t)j + h(t)k is smooth on an open interval I if f’, g’, and h’ are continuous on I and r’(t) for any value of t in the interval I.
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18. Parts formula
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