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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #66 : Equations Of Lines And Planes

Find the equation of the plane from the points on the plane p_1=(2,-5,6) , p_2=(3,-1,1) and p_3=(1,-4,1)

Note: Use the point p_1 when forming the equation of the plane

Possible Answers:

15x+10y+5z=50

-15x-10y+5z=-10

-15x+10y+5z=-30

-15x+10y+5z=-50

Correct answer:

-15x+10y+5z=-50

Explanation:

First, we need to form two vectors on the plane to get the normal vector to the plane. This is done from the following operation:

$v_1=p_1-p_2=\left \langle (2-3),(-5+1),(6-1) \right \rangle=\left \langle -1,-4,5\right \rangle$

$v_2=p_1-p_3=\left \langle (2-1),(-5+4),(6-1) \right \rangle=\left \langle 1,-1,5\right \rangle$

We then take the cross product of these vectors, which gets us the normal vector

$n=\left \langle -15,10,5\right \rangle$

Plugging the point p_1 and the normal vector into the equation of the plane, we get

-15(x-2)+10(y+5)+5(z-6)=0

Simplifying, we get

-15x+10y+5z=-50

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Example Question #67 : Equations Of Lines And Planes

Find the equation of the plane tangent to the surface z=f(x,y) at the point where $x=\frac{\pi}{2}$ and y = 2 .

$f(x,y)=\sin(2xy)$

Possible Answers:

$z=x-4\pi y-8\pi$

$z=2y\cos(2xy)-2x\cos(2xy)+\frac{\pi}{2}$

z=4x-4y-8

$z=4x+4y-8-2\pi$

$z=2x\cos(2xy)-2y\cos(2xy)+\frac{\pi-4}{2}$

Correct answer:

$z=4x+4y-8-2\pi$

Explanation:

Find the equation of the plane tangent to the surface z=f(x,y) at the point where $x=x_o=\frac{\pi}{2}$ and y = y_o=2 .

$f(x,y)=\sin(2xy)$ (1)

The equation of the plane tangent to the surface defined by f(x,y) is given by the formula:

z=f(x_o,y_o)+f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o) (2)

In Equation (2) f_x and f_y are the partial derivatives of with respect to and , respectively. For this particular problem we have $(x_o,y_o) =\left(\frac{\pi}{2},2\right)$

Let's fill in Equation (2) term-by-term:

$f(x_o,y_o)=f\left(\frac{\pi}{2},2\right)=\sin(2\pi)=0$

Compute the partial derivative and then evaluate both at $\left(\frac{\pi}{2}, 2 \right )$

- Partial with respect to x

$f_x(x,y)=\frac{\partial f}{\partial x}=2y\cos(2xy)$

$f_x\left(\frac{\pi}{2}, 2 \right )=2(2)\cos(2\pi)=4$

- Partial with respect to

$f_y(x,y)=2x\cos(2xy)$

$f_y\left(\frac{\pi}{2},2 \right )=2(2)\cos(2\pi) = 4$

Now fill in Equation (2) and simplify to get the equation of the tangent plane:

$z=4+4\left(x-\frac{\pi}{2}\right)+4(y-2)$

$=4x-2\pi+4y-8$

$=4x+4y-8-2\pi$

Therefore the equation of the tangent plane to the surface $z=\cos(2xy)$ at the point $\left(\frac{\pi}{2},2 \right )$ is simply

$z=4x+4y-8-2\pi$

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Example Question #61 : Equations Of Lines And Planes

Find the equation of the plane that contains the point (-5,-4,3) and the normal vector $n=\left \langle 2,1,-3\right \rangle$

Possible Answers:

2x+y-3z=-23

y+z=2

x-2y+3z=19

x+2y-3z=20

Correct answer:

2x+y-3z=-23

Explanation:

To find the equation of the plane, we use the formula A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where the point given is (x_0,y_0,z_0) and the normal vector $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 2(x+5)+1(y+4)-3(z-3)=0 . Manipulating the equation through algebra to isolate the variables, we get 2x+y-3z=-23 .

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Example Question #1 : Parametric Curves

Find the length of the parametric curve described by

$x = 2\sin t$

$y = 2\cos t$

from t=0 to $\pi$ .

Possible Answers:

None of the other answers

$\pi$

$4\pi$

$\frac{3\pi}{4}$

$2\pi$

Correct answer:

$2\pi$

Explanation:

There are several ways to solve this problem, but the most effective would be to notice that we can derive the following-

$x^2 = 4\sin^2t$

$y^2 = 4\cos^2t$

Hence

$x^2+y^2 = 4\sin^2t + 4\cos^2t = 4$

Therefore our curve is a circle of radius $\sqrt4=2$ , and it's circumfrence is $4\pi$ . But we are only interested in half that circumfrence ( is from to $\pi$ , not $2\pi$ .), so our answer is $2\pi$ .

Alternatively, we could've found the length using the formula

$l = \int_{t_0}^{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$ .

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Example Question #2 : Parametric Curves

Find the coordinates of the curve function

c(t)=<t^2,t,-3t>
when t=5 .

Possible Answers:

c(5)=<30,10,-5>

c(5)=<25,5,-15>

c(5)=<2,4,6>

c(5)=<15,10,-10>

Correct answer:

c(5)=<25,5,-15>

Explanation:

To find the coordinates, we set t=5 into the curve function.

We get

c(5)=<5^2,5,-3(5)>

and thus

c(5)=<25,5,-15>

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Example Question #181 : Calculus 3

Find the coordinates of the curve function

c(t)=<5,t^2,t>

when t=3

Possible Answers:

<5,9,3>

<0,0,3>

<0,3,0>

<-5,6,3>

Correct answer:

<5,9,3>

Explanation:

To find the coordinates, we evaluate the curve function for t=3

As such,

c(3)=<5, 3^2, 3)>=<5,9,3>

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Example Question #4 : Parametric Curves

Find the coordinates of the curve function

c(t)=<cos (t), sin (t), cos (t)>

when $t=\frac{\pi}{2}$

Possible Answers:

<0,1,0>

<1,0,1>

<1,1,1>

$<\frac{1}{2},-\frac{1}{2},\frac{1}{2}>$

Correct answer:

<0,1,0>

Explanation:

To find the coordinates, we evaluate the curve function for $t=\pi/2$

As such,

$c(\frac{\pi}{2})=<cos (\frac{\pi}{2}), sin (\frac{\pi}{2}), cos (\frac{\pi}{2})>=<0,1,0>$

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Example Question #5 : Parametric Curves

Find the equation of the line passing through the two points, given in parametric form:

(1, 2, 3), (0, -4, 1)

Possible Answers:

$\left \langle 1-t, 2+6t, 3+2t\right \rangle$

$\left \langle 1+t, 2+6t, 3+2t\right \rangle$

$\left \langle 1+t, 2-6t, 3+2t\right \rangle$

$\left \langle 1+t, 2+t, 3+t\right \rangle$

Correct answer:

$\left \langle 1+t, 2+6t, 3+2t\right \rangle$

Explanation:

To find the equation of the line passing through these two points, we must first find the vector between them:

$\vec{v}=\left \langle 1, 6, 2\right \rangle$

This was done by finding the difference between the x, y, and z components for the vectors. (This can be done in either order, it doesn't matter.)

Now, pick a point to be used in the equation of the line, as the initial point. We write the equation of line as follows:

$\vec{r}=\left \langle 1, 2, 3\right \rangle + t\left \langle 1, 6, 2\right \rangle=\left \langle 1+t, 2+6t, 3+2t\right \rangle$

The choice of initial point is arbitrary.

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Example Question #4 : Parametric Curves

Find the coordinate of the parametric curve when

$t=\pi/2$

$x=sin\,t$ , y=t^2

Possible Answers:

$(1/2, \pi/6)$

$(0,\pi/2)$

$(1,\pi^2/4)$

Correct answer:

$(1,\pi^2/4)$

Explanation:

To find the coordinates of the parametric curve we plug in for

$t=\pi/2$ .

$x=sin\,(\pi/2)= 1$

$y=(\pi/2)^2=\pi^2/4$

As such the coordinates are

$(1,\pi^2/4)$

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Example Question #2 : How To Find Velocity

The position of a particle is given by $s(t)=\frac{2t^2+1}{t+1}$ . Find the velocity at t=2 .

Possible Answers:

$\frac{15}{9}$

$\frac{16}{3}$

$\frac{15}{3}$

Correct answer:

$\frac{15}{9}$

Explanation:

The velocity is given as the derivative of the position function, or

$v(t)=\frac{d(s(t))}{dt}$ .

We can use the quotient rule to find the derivative of the position function and then evaluate that at t=2 . The quotient rule states that

$\left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$ .

In this case, $f(t)=2t^2+1\Rightarrow f'(t)=4t$ and $g(t)=t+1\Rightarrow g'(t)=1$ .

We can now substitute these values in to get

$\left[\frac{f(t)}{g(t)}\right]'=\frac{(t+1)*(4t)-(2t^2+1)*(1)}{(t+1)^2}=\frac{2t^2+4t-1}{(t+1)^2}$ .

Evalusting this at t=2 gives us $\frac{2*2^2+4*2-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}$ .

So the answer is $\frac{15}{9}$ .

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #66 : Equations Of Lines And Planes

Example Question #67 : Equations Of Lines And Planes

Example Question #61 : Equations Of Lines And Planes

Example Question #1 : Parametric Curves

Example Question #2 : Parametric Curves

Example Question #181 : Calculus 3

Example Question #4 : Parametric Curves

Example Question #5 : Parametric Curves

Example Question #4 : Parametric Curves

Example Question #2 : How To Find Velocity

All Calculus 3 Resources

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