 |
 |
 |
 |
|
|
|
|
1.7, #12. (a) Eliminate the parameter to
find a Cartesian equation of the curve.
(b) Sketch the curve and indicate with an arrow the direction in which
the curve is traced as the parameter increases.
 |
Here we use the meaning of the meaning of the natural logarithm to rewrite the relationship in exponential form. |
 |
Here we substitute the value for the parameter t into the second equation in place of t. |
 |
Taking a square root is equivalent to raising a quantity to the one-half power. |
 |
Simplifying using exponential laws. |
 |
Here is the plot of the equation:
In Maple, this would be: > plot(exp(x/2),x=1..5);
|
 |
Here is the same plot, using the parametric values from a table of values in Excel. |
 |
Here is the same plot in parametric form:
Using Maple this is: > plot([ln(x),sqrt(x),x=1..150]);
|
1.7, #25. Graph the curve: 
 |
> implicitplot(x=y-3*y^3+y^5,x=-3..3,y=-3..3,numpoints=10000); |
1.7, #30. If a projectile is fired with an initial velocity of 
meters per second at an angle 
above the horizontal and air resistance is assumed to be negligible, then its
position after t secons is given by the parametric equations: 
and 
where
g is the acceleration due to gravity 
(a) If a gun is fired with 
,
amd 
, when
will the bullet hit the ground? How far from the gun will the bullet
hit the ground? What is the maximum height reached by the bullet?
(b) Use a graphing device
to check your answers to part (a). Then graph the path of the projectile
for several other values of the angle
to see
where it hits the ground. Summarize your findings.
(c) Show that the path
is parabolic by eliminating the parameter.
 |
Substitute in the
and the . |
 |
Simplify |
 |
Set equal to zero because we are trying to find when the bullet has zero height, and substitute the value for gravity, and simplify. |
 |
Factor the right side. |
 |
t = 0 could indicate before the bullet was fired. We are interested in the t = 51.02 sec. |
 |
Part (b) parametric plot. Using Maple this would be:
> plot([250*sqrt(3)*t,250*t-4.9*t^2,t=0..52]); (Exact values for the sine and cosine of 30 degrees were used) |
 |
Here you will see that an angle of 20 degrees was used. To plot
correctly, a coversion is present to change to radians.
> plot([500*cos(20*3.14159/180)*t,500*
|
 |
Here you will see that an angle of 40 degrees was used. To plot
correctly, a coversion is present to change to radians.
> plot([500*cos(40*3.14159/180)*t,500*
|
If we can eliminate the parameter (c), then we can verify that this
is paraboloic.
 |
We substitute in the given values for
and ,
simplify, and solve for t. |
 |
Again, we substitute in the given values for
and .
Now we substitute our value for t which we found in the last step.
Then we simplify. The result is a paraboloa which opens downward. |
|
|
|
|
|
|
|
|
|
