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AP Calculus BC : Computation of Derivatives

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Example Question #4 : Derivatives Of Parametric, Polar, And Vector Functions

What is the derivative of $r=7sin\theta-6$ ?

Possible Answers:

$\frac{14cos\theta\sin\theta+6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta-6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7+14\sin^{2}\theta+6\sin\theta}$

$-\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

Correct answer:

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=7sin\theta-6$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=7cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$

$\frac{dy}{dx}=\frac{7cos\theta\sin\theta+(7sin\theta-6)\cos\theta}{7cos\theta\cos\theta-(7sin\theta-6)\sin\theta}$

$\frac{dy}{dx}=\frac{7cos\theta\sin\theta+7sin\theta\cos\theta-6\\cos\theta}{7cos\theta\cos\theta-7sin\theta\sin\theta+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(cos^{2}\theta-sin^{2}\theta)+6\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7((1-\sin^{2}\theta)-sin^{2}\theta)+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(1-2\sin^{2}\theta)+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

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Example Question #5 : Derivatives Of Parametric, Polar, And Vector Functions

Given that $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ . We define its gradient as :

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ be given by:

$f(x_{1},x_{2},\cdots, x_{n}})=\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

What is the gradient of ?

Possible Answers:

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{2\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{2x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{2x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+2x_{2}^2+\cdots+x_{n}^2}} \right]$

Correct answer:

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

Explanation:

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

$\frac{\partial f}{\partial x_{i}}=\frac{x_{i}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

To see this, fix all other variables and assume that you have only $x_{i}$ as the only variable.

Now we apply the given defintion , i.e,

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

with :

$\frac{\partial f}{\partial x_{1}}=\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{2}}=\frac{x_{2}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{3}}=\frac{x_{3}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\vdots$

$\frac{\partial f}{\partial x_{n}}=\frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

this gives us the solution .

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Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Let $f:\mathbb{R}^{n}\mapsto \mathbb{R}$ .

We define the gradient of as:

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f=sin(x_{1}+2x_{2}+\cdots nx_{n})$ .

Find the vector gradient.

Possible Answers:

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad -ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad cos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[x_{1}cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

Correct answer:

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

Explanation:

We note first that :

$\frac{\partial f}{\partial x_{1}}=cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{1}$ is the only variable here.

$\frac{\partial f}{\partial x_{2}}=2 cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{2}$ is the only variable here.

Continuing in this fashion we have:

$\frac{\partial f}{\partial x_{n}}=ncos(x_{1}+2x_{2}+\cdots nx_{n})$

Again using the Chain Rule and assuming that $x_{n}$ is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

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Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Let

$\vec{u}=(\sqrt{1+t^2},2\sqrt{1+t^2},3\sqrt{1+t^2},\cdots n\sqrt{1+t^2})$

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

Possible Answers:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{t}{\sqrt{t^2+1}},\cdots \frac{t}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{2t}{\sqrt{t^2+1}}\right)$

Correct answer:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

Explanation:

To find the derivative of this vector, all we need to do is to differentiate each component with respect to t.

Use the Power Rule and the Chain Rule when differentiating.

$(\sqrt{1+t^2})dt=(1+t^2)^\frac{1}{2}dt=\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{t}{\sqrt{t^2+1}}$ is the derivative of the first component.

$(2\sqrt{1+t^2})dt=2(1+t^2)^\frac{1}{2}dt=2\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{2t}{\sqrt{t^2+1}}$ of the second component.

$\vdots$

$(n\sqrt{1+t^2})dt=n(1+t^2)^\frac{1}{2}dt=n\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{nt}{\sqrt{t^2+1}}$ is the derivative of the last component . we obtain then:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

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Example Question #8 : Derivatives Of Parametric, Polar, And Vector Functions

$\newline {\vec{r}(t)} = (3e^t+t^3,e^{2t}+t,9e^t+e^{2t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,0,9e^t+2e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,2e^{2t},9e^t)$

Correct answer:

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

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:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$
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, ${\vec{r}(t)} = ((3e^t + t^3),(e^{2t} + t),(9e^t+e^{2t}))$

$\frac{\mathrm{d} }{\mathrm{d} t}(3e^t + t^3)= \frac{\mathrm{d} }{\mathrm{d} t}3e^t + \frac{\mathrm{d} }{\mathrm{d} t}t^3 = 3e^t + 3t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}(e^{2t} + t)= \frac{\mathrm{d} }{\mathrm{d} t}e^{2t}+ \frac{\mathrm{d} }{\mathrm{d} t}t =2e^{2t} + 1$

$\frac{\mathrm{d} }{\mathrm{d} t}(9e^t+e^{2t})= \frac{\mathrm{d} }{\mathrm{d} t}9e^t + \frac{\mathrm{d} }{\mathrm{d} t}e^{2t} = 9e^t + 2e^{2t}$

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

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

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Example Question #171 : Derivatives

Given the parametric curve

x= 2t^2 , $y=\sin{t}$

Evaluate $\frac{dy}{dx}$ when $t = \frac{\pi}{4}$ .

Possible Answers:

$\frac{\sqrt{2}}{2\pi}$

$\frac{2}{\pi}$

$\sqrt{2}$

$\frac{2\pi}{\sqrt{2}}$

Correct answer:

$\frac{\sqrt{2}}{2\pi}$

Explanation:

To find $\frac{dy}{dx}$ , we can use the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

$\frac{dy}{dt} = \frac{d}{dt}(\sin(t)) = \cos(t)$ .

And $\frac{dx}{dt} = \frac{d}{dt}(2t^2) = 4t$ .

Hence $\frac{dy}{dx} = \frac{\cos(t)}{4t}$ .

Plugging in $t = \frac{\pi}{4}$ , we get

$\frac{\cos(\pi/4)}{4\frac{\pi}{4}} = \frac{\sqrt{2}/2}{\pi} = \frac{\sqrt{2}}{2\pi}.$

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Example Question #71 : Computation Of Derivatives

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of the following function:

x=t^2e^t , y=t+3

Possible Answers:

2te^t+t^2e^t

$\frac{1}{e^t(t+2)}$

2te^t

Correct answer:

$\frac{1}{e^t(t+2)}$

Explanation:

To find the derivative of the parametric function, we must use the following:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}$

So, we must take the derivative of each component with respect to t:

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

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

$\frac{\mathrm{d} y}{\mathrm{d} t}=1$

This derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Dividing the two and factoring, we get

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{t}{2te^t+t^2e^t}=\frac{t(1)}{te^t(t+2)}=\frac{1}{e^t(t+2)}$

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Example Question #71 : Computation Of Derivatives

Find $\frac{\mathrm{d} x}{\mathrm{d} t}$ , where

x^2=t^3

Possible Answers:

$\frac{3t^2}{2x}$

$\frac{3x^2}{2t}$

$\frac{3t^2}{2}$

Correct answer:

$\frac{3t^2}{2x}$

Explanation:

To find the derivative of x with respect to t, we must differentiate both sides of the parametric equation with respect to t:

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

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Note that the chain rule was used when taking the derivative of x^2 .

Solving, we get

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

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Example Question #441 : Ap Calculus Bc

Screen shot 2016 03 30 at 4.58.19 pm

A particle moves around the xy plane such that its position as a function of time is given by the parametric function:

$\begin{pmatrix} x(t)\\y(t) \end{pmatrix}=\begin{pmatrix} \sin(3t)\\\sin(2t) \end{pmatrix}$ .

What is the slope, $\frac{dy}{dx}$ , of the particle's trajectory when $t=\pi/12$ ?

Possible Answers:

$\frac{\sqrt{6}}{3}$

$\frac{\sqrt{6}}{2}$

$\frac{\sqrt{2}}{2}$

$\frac{3\sqrt{2}}{2}$

$\frac{\sqrt{2}}{3}$

Correct answer:

$\frac{\sqrt{6}}{3}$

Explanation:

Evaluate the slope as

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ .

We have

$\frac{dy}{dt}=2\cos(2t)$

and

$\frac{dx}{dt}=3\cos(3t)$

$\frac{dy}{dx}=\frac{2\cos(2t)}{3\cos(3t)}$

Evaluating this when $t=\pi/12$ gives

$\frac{dy}{dx}=\frac{2\cos(\frac{\pi}{6})}{3\cos(\frac{\pi}{4})}=\frac{2\left (\frac{\sqrt{3}}{2} \right )}{3\left (\frac{1}{\sqrt{2}} \right )}=\frac{\sqrt{6}}{3}$ .

Remark: This curve is one example from family of curves called Lissajous figures, which can be observed on oscilloscopes.

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AP Calculus BC : Computation of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #4 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #5 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #8 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #171 : Derivatives

Example Question #71 : Computation Of Derivatives

Example Question #71 : Computation Of Derivatives

Example Question #441 : Ap Calculus Bc

All AP Calculus BC Resources

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