2.7 Cylindrical and spherical coordinates  (Page 9/11)

[T] The “bumpy sphere” with an equation in spherical coordinates is $\rho =a+b\text{cos}(m\theta )\text{sin}(n\phi ),$ with $\theta \in [0,2\pi ]$ and $\phi \in [0,\pi ],$ where $a$ and $b$ are positive numbers and $m$ and $n$ are positive integers, may be used in applied mathematics to model tumor growth.

1. Show that the “bumpy sphere” is contained inside a sphere of equation $\rho =a+b.$ Find the values of $\theta$ and $\phi$ at which the two surfaces intersect.
2. Use a CAS to graph the surface for $a=14,$ $b=2,$ $m=4,$ and $n=6$ along with sphere $\rho =a+b.$
3. Find the equation of the intersection curve of the surface at b. with the cone $\phi =\frac{\pi }{12}.$ Graph the intersection curve in the plane of intersection.

For vectors $\text{a}$ and $\text{b}$ and any given scalar $c,$ $c\left(\text{a}·\text{b}\right)=\left(c\text{a}\right)·\text{b}.$

For vectors $\text{a}$ and $\text{b}$ and any given scalar $c,$ $c\left(\text{a}\times \text{b}\right)=\left(c\text{a}\right)\times \text{b}.$

The symmetric equation for the line of intersection between two planes $x+y+z=2$ and $x+2y-4z=5$ is given by $-\frac{x-1}{6}=\frac{y-1}{5}=z.$

If $\text{a}·\text{b}=0,$ then $\text{a}$ is perpendicular to $\text{b}.$

$\text{a}=9\text{i}-2\text{j},\text{b}=\mathrm{-3}\text{i}+\text{j}$

1. $3\text{a}+\text{b}$
2. $\left|\text{a}\right|$
3. $\text{a}\times |\text{b}\times |\text{a}$
4. $\text{b}\times |\text{a}$

$\text{a}=2\text{i}+\text{j}-9\text{k},\text{b}=\text{−}\text{i}+2\text{k},\text{c}=4\text{i}-2\text{j}+\text{k}$

1. $2\text{a}-\text{b}$
2. $\left|\text{b}\times \text{c}\right|$
3. $\text{b}\times \left|\text{b}\times \text{c}\right|$
4. $\text{c}\times \left|\text{b}\times \text{a}\right|$
5. ${\text{proj}}_{\text{a}}\text{b}$

Find the values of $a$ such that vectors $⟨2,4,a⟩$ and $⟨0,\mathrm{-1},a⟩$ are orthogonal.

Find the unit vector that has the same direction as vector $\text{v}$ that begins at $\left(0,\mathrm{-3}\right)$ and ends at $\left(4,10\right).$

Find the unit vector that has the same direction as vector $\text{v}$ that begins at $\left(1,4,10\right)$ and ends at $\left(3,0,4\right).$

The parallelogram spanned by vectors $\text{a}=⟨1,13⟩\text{and}\text{b}=⟨3,21⟩$

The parallelepiped formed by $\text{a}=⟨1,4,1⟩\text{and}\text{b}=⟨3,6,2⟩,$ and $\text{c}=⟨\mathrm{-2},1,\mathrm{-5}⟩$

The line that passes through point $\left(2,\mathrm{-3},7\right)$ that is parallel to vector $⟨1,3,\mathrm{-2}⟩$

The line that passes through points $\left(1,3,5\right)$ and $\left(\mathrm{-2},6,\mathrm{-3}\right)$

The plane that passes through point $\left(4,7,\mathrm{-1}\right)$ and has normal vector $\text{n}=⟨3,4,2⟩$

The plane that passes through points $\left(0,1,5\right),\left(2,\mathrm{-1},6\right),\text{and}\left(3,2,5\right).$

$9x^2+4y^2-16y+36z^2=20$

$x^2=y^2+z^2$

$x^2+y^2+z^2=144$

$z=x^2+y^2-1$

${\rho }^2\left({\text{sin}}^2\left(\phi \right)-{\text{cos}}^2\left(\phi \right)\right)=1$

$r^2-2r\text{cos}\left(\theta \right)+z^2=1$

If the boat velocity is $5$ km/h due north in still water and the water has a current of $2$ km/h due west (see the following figure), what is the velocity of the boat relative to shore? What is the angle $\theta$ that the boat is actually traveling?

When the boat reaches the shore, two ropes are thrown to people to help pull the boat ashore. One rope is at an angle of $25\text{°}$ and the other is at $35\text{°}.$ If the boat must be pulled straight and at a force of $500\text{N},$ find the magnitude of force for each rope (see the following figure).

An airplane is flying in the direction of 52° east of north with a speed of 450 mph. A strong wind has a bearing 33° east of north with a speed of 50 mph. What is the resultant ground speed and bearing of the airplane?

Calculate the work done by moving a particle from position $(1,2,0)$ to $(8,4,5)$ along a straight line with a force $\text{F}=2\text{i}+3\text{j}-\text{k}.$

Because your tire is flat, you are only able to apply your force at a $60\text{°}$ angle. What is the torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

Someone lends you a tire jack and you are now able to apply a 200-N force at an $80\text{°}$ angle. Is your resulting torque going to be more or less? What is the new resulting torque at the center of the bolt? Assume this force is not enough to loosen the bolt.