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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : How To Find Distance

Find the distance traveled by an object from t=2 to t=5 seconds if the velocity of the object, in m/s, is described by the following equation:

$v(t)=\frac{1}{4}t-\frac{1}{2}$

Possible Answers:

$\frac{5}{2}$

$\frac{1}{2}$

$\frac{25}{8}$

$\frac{9}{8}$

Correct answer:

$\frac{9}{8}$

Explanation:

In order to find the distance traveled by an object we need an equation for position. If velocity is the derivative of position, then we must integrate the given equation from t=2 to t=5 to find the total distance traveled by the object over that interval:

$\int_{2}^{5}\left(\frac{1}{4}t-\frac{1}{2}\right)dt=\left[\frac{1}{8}t^2-\frac{1}{2}t\right]_{2}^{5}=\left(\frac{25}{8}-\frac{5}{2}\right)-\left(\frac{1}{2}-1\right)=\frac{9}{8}$

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Example Question #1 : How To Find Distance

Find the midpoint of the line segment connecting the points (-2,7) and (6,-3) .

Possible Answers:

None of these answers are correct.

(-2,2)

(-2,4)

(2,2)

Correct answer:

(2,2)

Explanation:

To find the midpoint between the two points, the average of the coordinates can be taken:

$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right )$

Where (x_1,y_1)=(-2,7) and (x_2,y_2)=(6,-3)

Plugging in these values we find our midpoint.

$\left(\frac{-2+6}{2},\frac{7-3}{2}\right) = \left(\frac{4}{2},\frac{4}{2}\right) = (2,2)$

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Example Question #2 : How To Find Distance

If the acceleration of an object at time is given by a(t) = t^2+3t+12 , what is the displacement of this particle from t =1 to t=2 , if its initial velocity was .

Possible Answers:

24.75

23.4

18.9

22.55

Correct answer:

24.75

Explanation:

To find the displacement of the object, we can integrate the equation for acceleration twice.

Integrating the acceleration equation once will give:

$v(t) = \frac{t^{2+1}}{2+1}+\frac{3t^{1+1}}{1+1}+12\cdott+C = \frac{1}{3}t^3 + \frac{3}{2}t^2+12t+C$

Because the initial velocity = 2, we can set v(0) =2 . Therefore,

$v(0) = \frac{1}{3}(0)^3+\frac{3}{2}(0)^2+12(0)+C = 2$

Therefore, C = 2 .

We can now use this in the velocity equation to give

$v(t) = \frac{1}{3}t^3+\frac{3}{2}t^2+12t+2$

integrating the velocity equation from t = 1 to t = 2 will give the distance

$\int_{5}^{7}\left[\frac{1}{3}t^3+\frac{3}{2}t^2+12t+2\right]dt \rightarrow$

$\rightarrow \frac{(2)^4}{12}+\frac{(2)^3}{2}+6(2)^2+2(2)+C- \frac{(1)^4}{12}-\frac{(1)^3}{2}-6(1)^2-2(1)-C$

$\rightarrow \frac{16}{12}+\frac{8}{2}+24+4- \frac{1}{12}-\frac{1}{2}-6-2 = 24.75$

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Example Question #3 : How To Find Distance

The velocity of a particle at time is given by the equation v(t) = 10t-7 . How far does this particle travel from t =2 to t = 6 ?

Possible Answers:

132

Correct answer:

132

Explanation:

To find the distance travelled, we must integrate the velocity equation

$x(t) = \int_{2}^{6}10t-7dt = 5(6)^2 - 7(6)-5(2)^2+7(2)$

x(t) = 180 - 42 -20 +14 = 132

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Example Question #4 : How To Find Distance

Find the distance from the first point to the second point.

First point: (0,4)

Second point: (2,6)

Possible Answers:

$distance=2\sqrt{3}$

$distance=\sqrt{3}$

distance=2

$distance=2\sqrt{2}$

$distance=\sqrt{2}$

Correct answer:

$distance=2\sqrt{2}$

Explanation:

To find the distance between two points we need to find the difference of the x-coordinates and y-coordinates between the two points.

So the x-coordinate is

2-0=2

The y-coordinate is

6-4=2

So the x-coordinate and y-coordinate is

(2,2)

The distance can be calculated by

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

So plugging in

$distance=\sqrt{2^2+2^2}$

$distance=\sqrt{4+4}$

$distance=\sqrt{8}$

$distance=2\sqrt{2}$

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Example Question #10 : How To Find Distance

Assume that two students are each walking on a circular path. We further suppose that these paths are centered at the origin. The radius of the first one is , and the radius of the second is . One student stopped walking and is located at (0,2) . What is the distance between the two students if other one is walking on the upper half of the circle of radius ?

Possible Answers:

$\sqrt{x^2+5-4\sqrt{1-x^2}}$

$\sqrt{2x^2-4\sqrt{1-x^2}}$

$\sqrt{2x^2+5\sqrt{1-x^2}}$

$\sqrt{2x^2+5-4\sqrt{1-x^2}}$

$\sqrt{5-4\sqrt{1-x^2}}$

Correct answer:

$\sqrt{5-4\sqrt{1-x^2}}$

Explanation:

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between this two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ . We will call the first fixed point (0,2) . To find the second point, note that the equation of the circle with radius 1 is given by:

x^2+y^2=1 . Since we are looking for the upper half, we have by solving for y:

$y=\sqrt{1-x^2}$ ( We need only the upper half in this case , that is why y is $\ge 0$ .

Hence our second point is $(x,y)=(x,\sqrt{1-x^2})$ .

Using the distance formula we have: $\sqrt{(x-0)^2+(\sqrt{1-x^2}-2)^2)}$ .

We need to simplify this expression a bit:

$=\sqrt{x^2+4 -4\sqrt{1-x^2}+1-x^2}$

this gives finally after cancelling and adding:

$\sqrt{5-4\sqrt{1-x^2}$

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Example Question #741 : Calculus

Possible Answers:

$\sqrt{5+4\sqrt{1-x^2}$

$\sqrt{5-4x\sqrt{1-x^2}$

$\sqrt{5+4\sqrt{1+x^2}$

$\sqrt{5-4\sqrt{1+x^2}$

$\sqrt{5-4x^2\sqrt{1-x^2}$

Correct answer:

$\sqrt{5+4\sqrt{1-x^2}$

Explanation:

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between this two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ . We will call the first fixed point (0,2) . To find the second point, note that the equation of the circle with radius 1 is given by:

x^2+y^2=1 .

Since we are looking for the lower half, we have by solving for y:

$y=-\sqrt{1-x^2}$ ( We need only the lower half in this case , that is why y is $\le 0$ .

Hence our second point is $(x,y)=(x,-\sqrt{1-x^2})$ .

Using the distance formula we have: $\sqrt{(x-0)^2+(\sqrt{1-x^2}+2)^2)}$ .

We need to simplify this expression a bit:

$=\sqrt{x^2+4 +4\sqrt{1-x^2}+1-x^2}$

this gives finally after cancelling and adding:

$\sqrt{+4\sqrt{1-x^2}$

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Example Question #742 : Calculus

A person is sitting on (1,1) and the other is walking on the line y=4 . What is the distance between the two?

Possible Answers:

$\sqrt{x^2-2x+10}$

$\sqrt{x^2+9}$

$\sqrt{(x-1)^2+3}$

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

$\sqrt{x+2}$

Correct answer:

$\sqrt{x^2-2x+10}$

Explanation:

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between these two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ .

We will call the first fixed point (1,1) .

To find the second point, note that the coordinate of the walking person is (x,4) since the x is changing and y is always 4.

Using the distance formula we have: $\sqrt{(x-1)^2+(4-1)^2}$ .

We need to simplify this expression a bit:

$\sqrt{(x^2-2x+1+9}$

this gives finally after cancelling and adding:

$\sqrt{x^2-2x+10}$

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Example Question #13 : How To Find Distance

Joe is sitting on (0,7) , and James is walking along the line y=0 .

What is the distance between the two ?( As a function of )

Possible Answers:

x+7

x^2+7

$\sqrt{(x-1)^2+7)}$

$\sqrt{x^2+49}$

$\sqrt{(x-7)^2+1}$

Correct answer:

$\sqrt{x^2+49}$

Explanation:

We will use the distance formula to show this result. Recall that having two points with coordinates $(x_{1},y_{1}),(x_{2},y_{2})$ , then the distance between these two points is given by:

$\sqrt {(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ .

We will call the first fixed point (0,7) . To find the second point, note that the coordinates of the walking person is (x,0) since James is walking along the x-axis.

Using the distance formula we have: $\sqrt{(x-0)^2+(7-0)^2}$ .

We need to simplify this expression a bit to get:

$\sqrt{x^2+49}$

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Example Question #14 : How To Find Distance

A person is restricted to sit on the center of the circle:

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

Another person is moving along the line y=4 . What is the distance between the two at any given position of the second?

Possible Answers:

$\sqrt{x^2+2x+10}$

$\sqrt{x^2-x+9}$

$\sqrt{x^2-2x+1}$

$\sqrt{x^2-2x+10}$

$\sqrt{x^2-2x+11}$

Correct answer:

$\sqrt{x^2-2x+10}$

Explanation:

To find the center of the circle. We will have to rewrite the expression of the circle.

We have x^2+y^2-2x-2y=2 .

Using complete the square method, this gives:

x^2-2x+1+y^2-2y+1=2+2=4 and we know that:

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

Therefore we have :

x^2+y^2-2x-2y=2 is the same as : (x-1)^2+(y-1)^2=4 .

This means that the center is (1,1). Since the moving has (x,4) as coordinates (the y-coordinate is the same, x is changing).

Using the distance formula we have :

$\sqrt{(x-1)^2 +(4-1)^2}$ .

We can simplify this expression to get :

$\sqrt{x^2-2x+1+9}=\sqrt{x^2-2x+10}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : How To Find Distance

Example Question #1 : How To Find Distance

Example Question #2 : How To Find Distance

Example Question #3 : How To Find Distance

Example Question #4 : How To Find Distance

Example Question #10 : How To Find Distance

Example Question #741 : Calculus

Example Question #742 : Calculus

Example Question #13 : How To Find Distance

Example Question #14 : How To Find Distance

All Calculus 1 Resources

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