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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #12 : Derivatives Of Vectors

Let $\vec{u}=(sin(2t),cos(2t),t^3)$ .

What is the third derivative of $\vec{u}$ ?

Possible Answers:

(-8cos(2x),8sin(2x),t)

(-8cos(2x),8sin(2x),3)

(-8cos(x),8sin(x),3t)

(-8cos(2x),8sin(2x),6)

(-8cos(2x),8sin(2x),2t^2)

Correct answer:

(-8cos(2x),8sin(2x),6)

Explanation:

To obtain the derivative, we simply differentiate 3 times each of the components.

Using the Chain Rule and the Power Rule we have :

$f(x)=sin(2t)\rightarrow f'(x)=2cos(2t)\rightarrow f''(x)=-4sin(2t)$

$\rightarrow cos(2x)^{[3]}=8sin(2x)$

$g(x)=cos(2t)\rightarrow g'(x)=-2sin(2t)\rightarrow g''(x)=-4cos(2t)$

$\rightarrow sin(2x)^{[3]}=-8cos(2x)$

$h(x)=t^3 \rightarrow h'(x)=3t^2 \rightarrow h''(x)=6t$

$\rightarrow t^3^{[3]}=6$

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Example Question #13 : Derivatives Of Vectors

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2,3,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3t,5t)$

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

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3,5)$

Explanation:

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

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

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

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Example Question #14 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (4t,8t,10t^{3})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (0,8,10t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,10t^{2})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t^{2})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t^{2})$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}4t,\frac{\mathrm{d} }{\mathrm{d} t}8t,\frac{\mathrm{d} }{\mathrm{d} t}10t^{3}\right) = (4,8,30t^{2})$

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Example Question #15 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (4t^{4},9t^{2},4t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (24,9t,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16t^3,9t,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (12,9,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16^{3},18t,4)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16^{3},18t,4)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1} \newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t}4t^4 = 4*4*t^{4-1} = 16*t^3$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}4t^{4},\frac{\mathrm{d} }{\mathrm{d} t}9t^{2},\frac{\mathrm{d} }{\mathrm{d} t}4t\right) = (16t^{3},18t,4)$

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Example Question #16 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (11t,6t,3t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,0,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,0)$

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,3)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,3)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1} \newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t}11t = 11*1*t^{1-1} = 11$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}11t,\frac{\mathrm{d} }{\mathrm{d} t}6t,\frac{\mathrm{d} }{\mathrm{d} t}3t\right) = (11,6,3)$

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Example Question #17 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (3t^3,10t+1,t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,10,1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^3,11,1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,0,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t,10,1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,10,1)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \ \frac{\mathrm{d} }{\mathrm{d} t}3t^3 = 3*3*t^{3-1} = 9*t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}3t^3,\frac{\mathrm{d} }{\mathrm{d} t}(10t+1),\frac{\mathrm{d} }{\mathrm{d} t}t\right) = (9t^2,10,1)$

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Example Question #18 : Derivatives Of Vectors

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

Possible Answers:

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2,4,6)$

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

Correct answer:

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

Explanation:

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

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

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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:

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

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Example Question #1002 : Calculus Ii

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

Possible Answers:

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

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

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

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

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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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #12 : Derivatives Of Vectors

Example Question #13 : Derivatives Of Vectors

Example Question #14 : Derivatives Of Vectors

Example Question #15 : Derivatives Of Vectors

Example Question #16 : Derivatives Of Vectors

Example Question #17 : Derivatives Of Vectors

Example Question #18 : Derivatives Of Vectors

Example Question #1001 : Calculus Ii

Example Question #1002 : Calculus Ii

Example Question #1003 : Calculus Ii

All Calculus 2 Resources

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