Derivatives of Parametric, Polar, and Vector Functions

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AP Calculus BC › Derivatives of Parametric, Polar, and Vector Functions

Questions 1 - 10

Solve for $\frac{dy}{dx}$ if and .

$\frac{3}{2}$

$-\frac{3}{2}$

$\frac{2}{3}$

$-\frac{2}{3}$

None of the above

Explanation

We can determine that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms will cancel out in the division process.

Since and , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all to derive

$\frac{dx}{dt}=(1)2t^{1-1}-(0)5=2$ and $\frac{dy}{dt}=(1)3t^{1-1}+(0)8=3$ .

Thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}$ .

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

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

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

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

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

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

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:

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

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

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

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 $\vec{u}$ ?

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

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

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

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

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

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:

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

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

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

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 ?

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

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

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

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

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

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,

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

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 .

Find the derivative $\frac{dy}{dx}$ of the polar function $r(\theta)=3-4sin(\theta)$ .

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=4cos(\theta)$

$\frac{dy}{dx}=-4cos(\theta)$

$\frac{dy}{dx}=\frac{-4cos(\theta)cos(\theta)+(3-4sin(\theta))sin(\theta)}{-4cos(\theta)sin(\theta)-(3-4sin(\theta))cos(\theta)}$

Explanation

The derivative of a polar function is found using the formula

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$

The only unknown piece is $r'(\theta)$ . Recall that the derivative of a constant is zero, and that

$\frac{d}{d\theta}sin(\theta)=cos(\theta)$ , so

$r'(\theta)=-4cos(\theta)$

Substiting $r'(\theta)\ and\ r(\theta)$ this into the derivative formula, we find

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos(\theta)cos(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

Find the derivative of the following polar equation:

$r=3+8sin\theta$

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

$\frac{8cos\theta sin\theta-6cos\theta }{4cos^2\theta+6sin\theta+3sin^2\theta}$

$\frac{8cos\theta sin\theta+3sin\theta }{4sin^2\theta-3sin\theta-8cos^2\theta}$

$\frac{16sin^2\theta+3cos\theta }{8cos^2\theta-3cos\theta-8sin^2\theta}$

$\frac{8cos\theta sin^2\theta+3cos^2\theta }{4cos^2\theta+3sin\theta-6sin^2\theta}$

Explanation

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to $\theta$ . This gives us:

$\frac{dr}{d\theta}=8cos\theta$

Now that we know dr/d $\theta$ , we can plug this value into the equation for the derivative of an expression in polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }$

Simplifying the equation, we get our final answer for the derivative of r:

$\frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }$

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 $t=\pi/12$ ?

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

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

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

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

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

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.

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

We define the gradient of as:

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

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

Find the vector gradient.

$[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})\]$

$[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})\]$

$[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})\]$

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})\]$

Find the derivative of the following set of parametric equations:

$\frac{2(7t-2)}{6t^2+10t-11}$

$\frac{-14t+1}{8t^2+5t-2}$

$\frac{4(3t+4)}{2t^2+4t-3}$

$\frac{t^2-7t+8}{6t-2}$

$\frac{5t^2+6t+3}{2t-11}$

Explanation

We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

$\frac{dx}{dt}=6t^2+10t-11$

$\frac{dy}{dt}=14t-4$

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

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

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}$

Find the first derivative of the polar function

$r(\theta)=sin(3\theta)$ .

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

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

$\frac{dy}{dx}=\frac{3sin(3\theta)cos(\theta)+cos(3\theta)sin(\theta)}{3sin(3\theta)sin(\theta)-cos(3\theta)cos(\theta)}$

$\frac{dy}{dx}=\frac{sin(3\theta)sin(\theta)+3cos(3\theta)cos(\theta)}{sin(3\theta)cos(\theta)-3cos(3\theta)sin(\theta)}$

Explanation

In general, the dervative of a function in polar coordinates can be written as

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$ .

Therefore, we need to find $r'(\theta)$ , and then substitute $r(\theta)\ and\ r'(\theta)$ into the derivative formula.

To find $r'(\theta)$ , the chain rule,

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ , is necessary.

We also need to know that

$\frac{d}{dx}sin(x)=cos(x)$ .

Therefore,

$r'(\theta)=3cos(3\theta)$ .

Substituting $r(\theta)\ and\ r'(\theta)$ into the derivative formula yields

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

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