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

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Example Questions

Example Question #1 : Derivatives Of Parametrics

Find the derivative of the following set of parametric equations:

x=2t^3+5t^2-11t+3

y=7t^2-4t-18

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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Example Question #2 : Derivatives Of Parametrics

Solve for $\frac{dy}{dx}$ if x=2t-5 and y=3t+8 .

Possible Answers:

$\frac{2}{3}$

$-\frac{3}{2}$

None of the above

$\frac{3}{2}$

$-\frac{2}{3}$

Correct answer:

$\frac{3}{2}$

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 x=2t-5 and y=3t+8 , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to derive

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

Thus:

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

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

Find the derivative of the following polar equation:

$r=3+8sin\theta$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #2 : Derivatives Of Polar Form

Find the first derivative of the polar function

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

Possible Answers:

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

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

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

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

Correct answer:

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

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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Example Question #441 : Ap Calculus Bc

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 #2 : 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:

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

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

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

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

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

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 #3 : 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})\]$

$[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})\]$

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 Vectors

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:

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

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

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

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

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

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 #442 : Ap Calculus Bc

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

Possible Answers:

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

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

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

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

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

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Derivatives Of Parametrics

Example Question #2 : Derivatives Of Parametrics

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

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #2 : Derivatives Of Polar Form

Example Question #441 : Ap Calculus Bc

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #3 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Vectors

Example Question #442 : Ap Calculus Bc

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