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Calculus 1 : Calculus

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Example Questions

Example Question #11 : Distance

We want to measure the distance of the intersection of the two circles:

x^2+y^2=1 and (x-1)^2+(y-1)^2=1 . What is this distance?

Possible Answers:

$1+\sqrt{2}}$

$\sqrt{5}$

$\sqrt{3}$

${}\sqrt{2}$

Correct answer:

${}\sqrt{2}$

Explanation:

To find the intersection of the two circles, we will have to solve the simultaneous system:

$\left\{\begin{matrix} x^2+y^2=1\\ (x-1)^2+(y-1)^2=1 \end{matrix}\right.$ .

We first write the second part of the system in a simplified manner , we have then:

(x-1)^2+(y-1)^2=x^2-2x+1+y^2-2y+1=1 and this gives:

x^2+y^2-2x-2y+1=0 .

Now we have :

x^2+y^2=1 , then this gives 2-2x-2y=0 and hence:

1-x-y=0 and solving for y we have y=1-x .

We use the fact that x^2+y^2=1 and therefore we have:

x^2+y^2=x^2+(1-x)^2=2x^2-2x+1=1 .

This gives us

x^2-x=0 .

Hence

$x=0\quad x=1$ .

Finally we deduce the value of y by plugging in each x value into our original equation and solving for y . When we do this we get x=0, y=1 , and when x=1, y=0 .

Now we the points (0,1) and (1,0) .

Using the distance formula we have :

$\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}$

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Example Question #751 : Calculus

The line y+2x-1=0 intersects the unit circle at two points. What is the distance between these points?

Possible Answers:

$\frac{4\sqrt5}{5}$

$\sqrt{3}$

$\sqrt{\frac{3}{5}}$

$\sqrt{\frac{5}{7}}$

$\frac{\sqrt{3}}{7}$

Correct answer:

$\frac{4\sqrt5}{5}$

Explanation:

We will have to find these points. After that we will use the distance formula to find the distance between them. To find the intersection we will have to solve the following system:

$\left\{\begin{matrix} x^2+y^2=1\\ y+2x-1=0\end{matrix}\right.$ which we can write by solving for y in the second equation.

$\left\{\begin{matrix} x^2+y^2=1\\ y=1-2x \end{matrix}\right.$

Plugging in the value of y in the first system we have:

(1-2x)^2+x^2=1 . This gives:

1-4x+4x^2+x^2=1 and hence we have:

5x^2-4x=0 .

This gives x=0 or $x=\frac{ 4}{5}$ .

We deduce the value of y from:

y=1-2x .

If x=0, y=1 .

If $x=\frac{4}{5}, y=-\frac{3}{5}$

We have the two points of interesection: $(0,1);\left(\frac{4}{5},\frac{-3}{5}\right)$ . Now we use the distance formula to the distance :

$\sqrt{\left(\frac{4}{5}\right)^2+\left(\frac{8}{5}\right)^2}=\frac{\sqrt{80}}{5}=\frac{4\sqrt5}{5}$

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Example Question #17 : How To Find Distance

The line y=1 intersects the circle of center (1,1) and radius at in two points. Find the distance between these two points.

Possible Answers:

$\sqrt{7}$

${}\sqrt{8}$

Correct answer:

Explanation:

We first find the equation of the circle. we have the center is (1,1) and the radius is 4. This gives:

(x-1)^2+(y-1)^2=16 .

We have the equation of the line is y=1.

Now we need to solve the system:

$\left\{\begin{matrix} y=1\\ (x-1)^2+(y-1)^2=16\end{matrix}\right.$ .

Plugging the value of y=1 in the second equation we have:

(x-1)^2=16 . Solving for x, we get :

x=-3 or x=5 . Therefore we have the two points:

(-3,1) and (5,1) . Now we use the distance formula:

$\sqrt{(-3-5)^2+(1-1)^2}=\sqrt{8^2}=8$

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Example Question #11 : Distance

Let (x,y) be any point on the plane. What is the distance between this point and its symmetric with respect to the -axis?

Possible Answers:

$y\sqrt{2}$

2|y|

$x\sqrt{2}}$

Correct answer:

2|y|

Explanation:

Let a=(x,y) , then its symmetric with respect to the x-axis is given by:

b=(x,-y) . Now using the distance formula we have:

$d(a,b)=\sqrt{(x-x)^2+(y-(-y))^2}=\sqrt{4y^2}=2|y|$

Note that $\sqrt{y^2}=|y|$ .

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Example Question #12 : Distance

Let a=(x,y) be any point in the plane. What is the distance between and its symmetric with respect to the -axis?

Possible Answers:

$y\sqrt{2}$

$x\sqrt{2}$

2|x|

Correct answer:

2|x|

Explanation:

Note that if a=(x,y) then its symmetric with respect to the y-axis is given by:

b=(-x,y) .

Now we use the distance formula.

$d(a,b)=\sqrt{(-x-x)^2+(y-y)^2}=\sqrt{4x^2}=2|x|$ .

Please note that $\sqrt{x^2}=|x|$ . Therefore the distance is 2|x| .

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Example Question #20 : How To Find Distance

Let a=(x,y) be any point in the plane. What is the distanc between and its symmetric with respect to the origin?

Possible Answers:

$\sqrt{x^2+y^2}$

$\sqrt{x^2-y^2}$

$\sqrt{y^2-x^2}$

$2\sqrt{x^2+y^2}$

2x+2y

Correct answer:

$2\sqrt{x^2+y^2}$

Explanation:

Let a=(x,y) . We know that its symmetric with respect to the origin is given by:

b= (-x,-y) .

Now we will use the distance formula to find the distance between a and b.

We have:

$d(a,b)=\sqrt{(x-(-x))^2+(y-(-y))^2}=\sqrt{4x^2+4y^2}$

and this gives: $d(a,b)=\sqrt{4}\sqrt{x^2+y^2}=2\sqrt{x^2+y^2}$ .

This shows the required result.

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Example Question #21 : How To Find Distance

Let B=(2a-x,y) . Let A=(x,y) . What is the distance between these two points?

Possible Answers:

|x-y|

2|x+y|

$\sqrt{2}|x-a|$

2|a|

x+y

Correct answer:

$\sqrt{2}|x-a|$

Explanation:

Let A=(x,y) and (2a-x,y) .

We will use the distance formula to find the distance d(A,B).

By definition we have:

$D(A,B)=\sqrt{(2x-2a)^2+(y-y)^2}=|x-a|\sqrt{2}$

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Example Question #751 : Spatial Calculus

What is the average function value of y=x^2 from the interval [1,3] ?

Possible Answers:

$\frac{8}{3}$

$\frac{13}{3}$

Correct answer:

$\frac{13}{3}$

Explanation:

Write the formula for average function value.

$f_{average}= \frac{1}{b-a}\int_{a}^{b} f(x)dx$

Substitute the function and the bounds.

$f_{average}= \frac{1}{3-1}\int_{1}^{3}x^2dx$

$f_{average}= \left(\frac{1}{2}\right)\left[\frac{1}{3}x^3\right]_{1}^{3}=\left(\frac{1}{2}\right)\left[\frac{1}{3}(3)^3-\frac{1}{3}(1)^3\right]$

$f_{average}= \frac{1}{2}\left[9-\frac{1}{3}\right]=\frac{1}{2}\left[\frac{26}{3}\right]=\frac{13}{3}$

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Example Question #21 : Distance

An object is fired in the air, and the function which describes the object's height is y=-9x^2+8x+1 , where is in seconds. What is the height of the firing point to the maximum height of the object?

Possible Answers:

$\frac{25}{9}$

$\frac{16}{9}$

$\frac{5}{4}$

Correct answer:

$\frac{16}{9}$

Explanation:

Use the vertex formula to determine the location maximum height of the parabola.

$x=-\frac{b}{2a}=-\frac{8}{2(-9)}=\frac{4}{9}$

Substitute this value to the position function to find the height.

$y=-9\left(\frac{4}{9}\right)^2+8\left(\frac{4}{9}\right)+1$

$y=-9\left(\frac{16}{81}\right)+\frac{32}{9}+1$

$y=-\frac{16}{9}+\frac{32}{9}+\frac{9}{9}=\frac{25}{9}$

This is the maximum height, but at the start point x=0 , the y-intercept shows that the projectile is fired at a height of 1. Subtract the maximum from the height of the projectile.

$\frac{25}{9}-1=\frac{25}{9}-\frac{9}{9}= \frac{16}{9}$

The height from firing point to the maximum is $\frac{16}{9}$ .

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Example Question #22 : Distance

Find the midpoint of the line segment connecting the points (0,3,-4) and (2,-6,3) .

Possible Answers:

(1,2,1)

(0.5,-2,-1)

(2,-3,-1)

(1,-1.5,-0.5)

Correct answer:

(1,-1.5,-0.5)

Explanation:

The coordinates of the midpoint is the average of the coordinates of each point.

$\left(\frac{x_1+x_1}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substituting in the values given:

$\left(\frac{0+2}{2},\frac{3+(-6)}{2},\frac{-4+3}{2}\right)$

$\left(\frac{2}{2},\frac{-3}{2},\frac{-1}{2}\right)$

Therefore the coordinates of the midpoint are (1,-1.5,-0.5) .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : Distance

Example Question #751 : Calculus

Example Question #17 : How To Find Distance

Example Question #11 : Distance

Example Question #12 : Distance

Example Question #20 : How To Find Distance

Example Question #21 : How To Find Distance

Example Question #751 : Spatial Calculus

Example Question #21 : Distance

Example Question #22 : Distance

All Calculus 1 Resources

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