Derivative of a parametric parabola
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Find
directly from the parametric equations
and evaluate
at
First, let’s discuss this function. . Is the dependence of
x
on
t
linear, quadratic, cubic, or other?
Linear.
Is the dependence of
y
on
t
linear, quadratic, cubic, or other?
It is quadratic.
What can we then expect for the shape of the graph of
y
vs.
x
?
The dependence of
y
on
x
will be quadratic, so we should expect a parabola.
Try to draw the graph from the parametric equations without using your graphing calculator. Then check your work by clicking “Next”.
alt="Graph of a parabola. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
Next let’s determine the point associated with
. Determine the
x
-coordinate of that point?
Determine the
y
-coordinate of that point?
Show this point on your graph and show the tangent line. Then check by clicking “Next”.
alt="Graph of a parbola with a tangent line. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
What is the algebraic sign of the slope of the tangent line?
When a line slopes up to the right, it has a positive “rise” over a positive “run”. Its slope is positive.
What will be the sign of
at this point?
It also is positive.
Now let’s determine
from the parametric equations. State the general relationship among
.
Determine
Determine
Substitute these results in
.
Evaluate
.
Is this consistent with the graph?
Yes, the tangent line is steeper than a line of slope = 1.
Now let’s determine
from the parametric equations. State the general relationship between
.
Apply this to our result for
We get
.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
General Contents
Detailed Contents
Index