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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #221 : Spatial Calculus

The position of a particle is given by the functions $\displaystyle x(t)=t^3$, $y(t)=5^{\sqrt[3]{t}}$ $\displaystyle y(t)=5^{\sqrt[3]{t}}$. What is the velocity of the particle at time t=3 $\displaystyle t=3$ ?

Possible Answers:

$\displaystyle (27,1.13)$

$\displaystyle (9,9.12)$

$\displaystyle (27,9.65)$

$\displaystyle (27,2.63)$

$\displaystyle (27,3.89)$

Correct answer:

$\displaystyle (27,2.63)$

Explanation:

Velocity of a particle can be found by taking the derivative of the position function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of position with respect to time, we are evaluating how position changes over time; i.e velocity!

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position functions

$\displaystyle x(t)=t^3$, $y(t)=5^{\sqrt[3]{t}}$ $\displaystyle y(t)=5^{\sqrt[3]{t}}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity functions are

$\displaystyle v_x(t)=3t^2;v_x(3)=3(3^2)=27$

$v_y(t)=\frac{5^{\sqrt[3]t}}{3\sqrt[3]{t^2}}ln(5);v_y(3)=\frac{5^{\sqrt[3]{3}}}{3\sqrt[3]{3^2}}ln(5)=2.63$ $\displaystyle v_y(t)=\frac{5^{\sqrt[3]t}}{3\sqrt[3]{t^2}}ln(5);v_y(3)=\frac{5^{\sqrt[3]{3}}}{3\sqrt[3]{3^2}}ln(5)=2.63$

$\displaystyle (27,2.63)$

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Example Question #222 : Spatial Calculus

Find the velocity of a particle at time t=1.5 $\displaystyle t=1.5$ if its position is given by the function $p(t)=2.7^{e^{2t}}$ $\displaystyle p(t)=2.7^{e^{2t}}$

Possible Answers:

$1.77\cdot 10^{6}$ $\displaystyle 1.77\cdot 10^{6}$

$7.11\cdot 10^{12}$ $\displaystyle 7.11\cdot 10^{12}$

$3.92\cdot 10^{9}$ $\displaystyle 3.92\cdot 10^{9}$

$1.84\cdot 10^{10}$ $\displaystyle 1.84\cdot 10^{10}$

$9.82\cdot 10^{10}$ $\displaystyle 9.82\cdot 10^{10}$

Correct answer:

$1.84\cdot 10^{10}$ $\displaystyle 1.84\cdot 10^{10}$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=2.7^{e^{2t}}$ $\displaystyle p(t)=2.7^{e^{2t}}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[e^u]=e^udu$

$\displaystyle d[a^u]=a^uduln(a)$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2e^{2t}(2.7^{e^{2t}})ln(2.7)$ $\displaystyle v(t)=2e^{2t}(2.7^{e^{2t}})ln(2.7)$

$v(1.5)=2e^{2(1.5)}(2.7^{e^{2(1.5)}})ln(2.7)=1.84\cdot 10^{10}$ $\displaystyle v(1.5)=2e^{2(1.5)}(2.7^{e^{2(1.5)}})ln(2.7)=1.84\cdot 10^{10}$

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Example Question #223 : Spatial Calculus

The position of particle is given by the function $p(t)=t^2{2^{t}}$ $\displaystyle p(t)=t^2{2^{t}}$. What is the velocity of the particle at time t=3 $\displaystyle t=3$ ?

Possible Answers:

27.2 $\displaystyle 27.2$

1.9 $\displaystyle 1.9$

1034.5 $\displaystyle 1034.5$

97.9 $\displaystyle 97.9$

118.3 $\displaystyle 118.3$

Correct answer:

97.9 $\displaystyle 97.9$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=t^2{2^{t}}$ $\displaystyle p(t)=t^2{2^{t}}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Product rule: $\displaystyle d[uv]=udv+vdu$

Note that u and v may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2t{2^t}+t^2{2^t}ln(2)$ $\displaystyle v(t)=2t{2^t}+t^2{2^t}ln(2)$

$v(t)=t{2^{t+1}}+t^2{2^t}ln(2)$ $\displaystyle v(t)=t{2^{t+1}}+t^2{2^t}ln(2)$

$v(3)=(3){2^{3+1}}+3^2{2^3}ln(2)=97.9$ $\displaystyle v(3)=(3){2^{3+1}}+3^2{2^3}ln(2)=97.9$

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Example Question #224 : Spatial Calculus

Find the velocity of a particle if its position function is $p(t)=3^{t^2}t$ $\displaystyle p(t)=3^{t^2}t$

Possible Answers:

$3e^{t^2}+2t^23^{t^2-1}ln(3)$ $\displaystyle 3e^{t^2}+2t^23^{t^2-1}ln(3)$

$3e^{t^2}+2t^23^{t^2}$ $\displaystyle 3e^{t^2}+2t^23^{t^2}$

$3e^{t^2}+2t^23^{t^2}ln(3)$ $\displaystyle 3e^{t^2}+2t^23^{t^2}ln(3)$

$3e^{t^2}+2t^23^{t^2-1}$ $\displaystyle 3e^{t^2}+2t^23^{t^2-1}$

$3te^{t^2}+2t^33^{t^2}ln(3)$ $\displaystyle 3te^{t^2}+2t^33^{t^2}ln(3)$

Correct answer:

$3e^{t^2}+2t^23^{t^2}ln(3)$ $\displaystyle 3e^{t^2}+2t^23^{t^2}ln(3)$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=3^{t^2}t$ $\displaystyle p(t)=3^{t^2}t$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Product rule: $\displaystyle d[uv]=udv+vdu$

Note that u and v may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=3e^{t^2}+2t^23^{t^2}ln(3)$ $\displaystyle v(t)=3e^{t^2}+2t^23^{t^2}ln(3)$

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Example Question #225 : Spatial Calculus

The position of a particle is given by the function $p(t)=4^{2t}$ $\displaystyle p(t)=4^{2t}$. What is the velocity of the particle at time t=2 $\displaystyle t=2$ ?

Possible Answers:

512 $\displaystyle 512$

709.8 $\displaystyle 709.8$

256 $\displaystyle 256$

177.4 $\displaystyle 177.4$

189.3 $\displaystyle 189.3$

Correct answer:

709.8 $\displaystyle 709.8$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=4^{2t}$ $\displaystyle p(t)=4^{2t}$

We'll need to make use of the following derivative rule:

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2(4^{2t})ln(4)$ $\displaystyle v(t)=2(4^{2t})ln(4)$

$v(2)=2(4^{2(2)})ln(4)=709.8$ $\displaystyle v(2)=2(4^{2(2)})ln(4)=709.8$

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Example Question #226 : Spatial Calculus

The position of a particle is given by the function $p(t)=1.2t^{3.89}$ $\displaystyle p(t)=1.2t^{3.89}$. What is the particle's velocity at time t=1.7 $\displaystyle t=1.7$ ?

Possible Answers:

41.77 $\displaystyle 41.77$

21.63 $\displaystyle 21.63$

62.51 $\displaystyle 62.51$

17.18 $\displaystyle 17.18$

32.09 $\displaystyle 32.09$

Correct answer:

21.63 $\displaystyle 21.63$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

For the position function $p(t)=1.2t^{3.89}$ $\displaystyle p(t)=1.2t^{3.89}$, the velocity function is

$v(t)=(3.89)(1.2)(t^{2.89})$ $\displaystyle v(t)=(3.89)(1.2)(t^{2.89})$

$v(1.7)=(3.89)(1.2)(1.7^{2.89})=21.63$ $\displaystyle v(1.7)=(3.89)(1.2)(1.7^{2.89})=21.63$

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Example Question #227 : Spatial Calculus

The position of a particle is given by the function $p(t)=4t^2-\frac{1}{2}e^t$ $\displaystyle p(t)=4t^2-\frac{1}{2}e^t$. At what time is the particle first stationary?

Possible Answers:

0.07 $\displaystyle 0.07$

4.21 $\displaystyle 4.21$

2.19 $\displaystyle 2.19$

$\displaystyle 0$

0.13 $\displaystyle 0.13$

Correct answer:

0.07 $\displaystyle 0.07$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=4t^2-\frac{1}{2}e^t$ $\displaystyle p(t)=4t^2-\frac{1}{2}e^t$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[e^u]=e^udu$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=8t-\frac{1}{2}e^t$ $\displaystyle v(t)=8t-\frac{1}{2}e^t$

The particle is stationary when this velocity is zero:

$8t-\frac{1}{2}e^t=0$ $\displaystyle 8t-\frac{1}{2}e^t=0$

$\displaystyle t=0.07,4.21$

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Example Question #228 : Spatial Calculus

The position of a particle starting at t=0 $\displaystyle t=0$ is given by the function $\displaystyle p(t)=2t+cos(t^3)$. At what time is the particle first stationary?

Possible Answers:

-2.12 $\displaystyle -2.12$

0.94 $\displaystyle 0.94$

-2.32 $\displaystyle -2.32$

1.41 $\displaystyle 1.41$

-1.51 $\displaystyle -1.51$

Correct answer:

0.94 $\displaystyle 0.94$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$\displaystyle p(t)=2t+cos(t^3)$

We'll need to make use of the following derivative rule(s):

$\displaystyle d[cos(u)]=-sin(u)du$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is:

$\displaystyle v(t)=2-3t^2sin(t^3)$

The particle is stationary when this function equals zero:

$\displaystyle 2-3t^2sin(t^3)=0$

Though there are multiple roots to this equation:

$\displaystyle t=-2.32,-2.12,-1.83,-1.51,0.94,1.41,1.86,2.10$

However, for the time that the particle is first stationary, we'll only consider the first positive value:

t=0.94 $\displaystyle t=0.94$

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Example Question #229 : Spatial Calculus

The position of a particle starting at time t=0 $\displaystyle t=0$ is given by the function $\displaystyle p(t)=2^t-sin(2t)$. When is the particle first stationary?

Possible Answers:

-14.92 $\displaystyle -14.92$

-10.21 $\displaystyle -10.21$

0.52 $\displaystyle 0.52$

1.12 $\displaystyle 1.12$

0.36 $\displaystyle 0.36$

Correct answer:

0.52 $\displaystyle 0.52$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$\displaystyle p(t)=2^t-sin(2t)$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Trigonometric derivative:

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2^{t}ln(2)-2cos(2t)$ $\displaystyle v(t)=2^{t}ln(2)-2cos(2t)$

To find when the particle is stationary, set this equation equal to zero:

$2^{t}ln(2)-2cos(2t)=0$ $\displaystyle 2^{t}ln(2)-2cos(2t)=0$

There are multiple negative values which satisfy this equation; however, since we start at t=0 $\displaystyle t=0$, the answer is

t=0.52 $\displaystyle t=0.52$

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Example Question #230 : Spatial Calculus

The position of a particle is given by the function $p(t)=t^4-3^{t}$ $\displaystyle p(t)=t^4-3^{t}$. When is the particle first stationary?

Possible Answers:

6.13 $\displaystyle 6.13$

4.96 $\displaystyle 4.96$

3.85 $\displaystyle 3.85$

0.91 $\displaystyle 0.91$

0.17 $\displaystyle 0.17$

Correct answer:

0.91 $\displaystyle 0.91$

Explanation:

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

To take the derivative of the position function

$p(t)=t^4-3^{t}$ $\displaystyle p(t)=t^4-3^{t}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$\displaystyle d[a^u]=a^uduln(a)$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=4t^3-3^{t}ln(3)$ $\displaystyle v(t)=4t^3-3^{t}ln(3)$

To find when the particle is stationary, set this value equal to zero:

$4t^3-3^{t}ln(3)=0$ $\displaystyle 4t^3-3^{t}ln(3)=0$

$\displaystyle t=0.91, 6.13$

The particle is first stationary at time t=0.91 $\displaystyle t=0.91$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #221 : Spatial Calculus

Example Question #222 : Spatial Calculus

Example Question #223 : Spatial Calculus

Example Question #224 : Spatial Calculus

Example Question #225 : Spatial Calculus

Example Question #226 : Spatial Calculus

Example Question #227 : Spatial Calculus

Example Question #228 : Spatial Calculus

Example Question #229 : Spatial Calculus

Example Question #230 : Spatial Calculus

All Calculus 1 Resources

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