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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1001 : Calculus Ii

$\newline {\vec{r}(t)} = (t^4+1,t^2+t,4t+3)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4t,2t,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4t^3,2t+1,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4t^3,2,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4t^3,2t,0)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4t^3,2t+1,4)$

Explanation:

$\newline \textnormal{In general:}\newline \textnormal{If} \ {\vec{r}(t)} = (f(t),g(t),z(t))\newline \textnormal{Then} \ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t)\right)$

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^4 + 1) = \frac{\mathrm{d} }{\mathrm{d} t}t^4 + \frac{\mathrm{d} }{\mathrm{d} t}1 = 4*t^{4-1} + 0 = 4*t^3$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}(t^4+1),\frac{\mathrm{d} }{\mathrm{d} t}(t^2+t),\frac{\mathrm{d} }{\mathrm{d} t}(4t+3)\right) = (4t^3,2t+1,4)$

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Example Question #482 : Parametric, Polar, And Vector

$\newline {\vec{r}(t)} = (cos(3t)+1,t^2+t,3t+11)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-3sin(3t),2,3)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3cos(3t),2t,3)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-3sin(3t),2t,3)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-3sin(3t),2t+1,3)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-3sin(3t),2t+1,3)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \frac{\mathrm{d} }{\mathrm{d} t}(cos(3t) + 1) = \frac{\mathrm{d} }{\mathrm{d} t}cos(3t) + \frac{\mathrm{d} }{\mathrm{d} t}1 = -3sin(3t)$

$\\ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}(cos(3t)+1),\frac{\mathrm{d} }{\mathrm{d} t}(t^2+t),\frac{\mathrm{d} }{\mathrm{d} t}(3t+11)\right) = (-3sin(3t),2t+1,3)$

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Example Question #1003 : Calculus Ii

$\newline {\vec{r}(t)} = (cos(4t),sin(t),sin(2t)+t^2)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),cos(t),2cos(2t)+2t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),sin(t),sin(2t)+t^2)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (cos(4t),sin(t),sin(2t)+2t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4sin(4t),sin(t),2sin(2t))$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),cos(t),2cos(2t)+2t)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \frac{\mathrm{d} }{\mathrm{d} t}(sin(2t) + t^2) = \frac{\mathrm{d} }{\mathrm{d} t}sin(2t) + \frac{\mathrm{d} }{\mathrm{d} t}t^2 = 2*cos(2t) + 2t$

$\\ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}cos(4t),\frac{\mathrm{d} }{\mathrm{d} t}sin(t),\frac{\mathrm{d} }{\mathrm{d} t}(sin(2t)+t^2)\right) \\ \\= (-4sin(4t),cos(t),2cos(2t)+2t)$

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Example Question #1004 : Calculus Ii

$\newline {\vec{r}(t)} = (sin(5t)+3t,sin(t^2),cos(t))\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tcos(t^2),-sin(t))$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,cos(t^2),-sin(t))$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5sin(5t),2tsin(t^2),sin(t))$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tsin(t^2),-sin(t))$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tcos(t^2),-sin(t))$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \frac{\mathrm{d} }{\mathrm{d} t}(sin(5t)+3t) = \frac{\mathrm{d} }{\mathrm{d} t}(5t)*cos(5t) + \frac{\mathrm{d} }{\mathrm{d} t}3t = 5*cos(5t) + 3$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}(sin(5t)+3t),\frac{\mathrm{d} }{\mathrm{d} t}sin(t^2),\frac{\mathrm{d} }{\mathrm{d} t}cos(t)\right) = (5cos(5t)+3,2tcos(t^2),-sin(t))$

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Example Question #1005 : Calculus Ii

$\newline {\vec{r}(t)} = (t^3+t,t^2+5,6t^3+t^2+t+1)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2t,18^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,2t,18^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,2t,18t^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2,18^2+2t+1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2t,18^2+2t+1)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \frac{\mathrm{d} }{\mathrm{d} t}(t^3+t) = \frac{\mathrm{d} }{\mathrm{d} t}t^3 + \frac{\mathrm{d} }{\mathrm{d} t}t = 3t^2 + 1$

$\\ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}(t^3+t),\frac{\mathrm{d} }{\mathrm{d} t}(t^2+5),\frac{\mathrm{d} }{\mathrm{d} t}(6t^3+t^2+t+1)\right)\\ \\ = (3t^2+1,2t,18t^2+2t+1)$

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Example Question #141 : Vector

$\newline {\vec{r}(t)} = (t^2,5t^3+t^2,e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15t^2,5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15t^2+t^2,5e^{5t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,5e^{5t})$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \frac{\mathrm{d} }{\mathrm{d} t}e^{5t} = \frac{\mathrm{d} }{\mathrm{d} t}(5t) * e^{5t} = 5e^{5t}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}t^2,\frac{\mathrm{d} }{\mathrm{d} t}(5t^3+t^2),\frac{\mathrm{d} }{\mathrm{d} t}e^{5t}\right) = (2t,15t^2+2t,5e^{5t})$

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Example Question #1003 : Calculus Ii

$\newline {\vec{r}(t)} = (20t,6t,3t^2)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (0,0,6t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,6)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,6t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,3)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,6t)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

In this problem, ${\vec{r}(t)} = (20t,6t,3t^2)$

$\frac{\mathrm{d} }{\mathrm{d} t}20t=20$

$\frac{\mathrm{d} }{\mathrm{d} t}6t=6$

$\frac{\mathrm{d} }{\mathrm{d} t}3t^2=6t$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,6t)$

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Example Question #142 : Vector

$\newline {\vec{r}(t)} = (e^{2t},e^{3t},11e^{2t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^{2t},e^{3t},e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2,3,22)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (e^{2t},e^{3t},11e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{2t}=2e^{2t}$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{3t}=3e^{3t}$

$\frac{\mathrm{d} }{\mathrm{d} t}11e^{2t}=22e^{2t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

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Example Question #21 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (e^{11t},e^{8t},e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},e^{8t},5e^{t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,8,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{4t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (e^{11t},e^{8t},e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{11t}=11e^{11t}$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{8t}=8e^{8t}$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{5t}=5e^{5t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

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Example Question #1001 : Calculus Ii

$\newline {\vec{r}(t)} = (5t^3,e^{11t},e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,e^{11t},5e^{5t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (5t^3,e^{11t},e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}5t^3=15t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{11t}=11e^{11t}$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{5t}=5e^{5t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11e^{11t},5e^{5t})$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1001 : Calculus Ii

Example Question #482 : Parametric, Polar, And Vector

Example Question #1003 : Calculus Ii

Example Question #1004 : Calculus Ii

Example Question #1005 : Calculus Ii

Example Question #141 : Vector

Example Question #1003 : Calculus Ii

Example Question #142 : Vector

Example Question #21 : Derivatives Of Vectors

Example Question #1001 : Calculus Ii

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