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17Calculus Vector Functions - Projectile Motion

17Calculus

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This page covers the basics of working with the position, velocity and acceleration vector functions. The acceleration vector that you learn about on this page can be expressed in terms of the unit tangent vector and the principal unit normal vector, which you can find on the acceleration vector components page.

Calculating The Velocity and Acceleration

Projectile motion using vector functions works just as you would expect. The following table lists the equations. If \(\vec{r}(t)\) is a vector function describing the position of a projectile, the velocity is \(\vec{v}(t)=\vec{r}'(t)\) and the acceleration is \(\vec{a}(t)=\vec{v}'(t)=\vec{r}''(t)\).

Position

\(\vec{r}(t) =x(t)\vhat{i}+y(t)\vhat{j}+z(t)\vhat{k}\)

Velocity

\(\vec{v}(t)=x'(t)\vhat{i}+y'(t)\vhat{j}+z'(t)\vhat{k}\) \(=\vec{r}'(t) \)

Speed

\(\| \vec{v}(t) \|\)

Acceleration

\(\vec{a}(t)=x''(t)\vhat{i}+y''(t)\vhat{j}+z''(t)\vhat{k}\) \(=\vec{v}'(t)\) \(=\vec{r}''(t)\)

There are really no surprises here. Notice that the speed is just the magnitude of the velocity and so it's value is always a positive scalar. Some instructors use the terms speed and velocity interchangeably but they actually refer to different things.

Calculating The Position Vector

Sometimes we are given the acceleration vector or the velocity vector and asked to calculate the position vector. In those cases we use integration. As you would expect, to get the velocity vector from the acceleration vector, we use these equations.

Acceleration

\(\vec{a}(t)=a_x(t)\vhat{i}+a_y(t)\vhat{j}+a_z(t)\vhat{k}\)

Velocity

\(\vec{v}(t)=\int{a_x(t)~dt}\vhat{i} + \int{a_y(t)~dt}\vhat{j} + \int{a_z(t)~dt}\vhat{k} + \vec{C} =\) \(\int{\vec{a}(t)~dt} + \vec{C}\)

Note - - In the equation for the velocity vector, the vector \(\vec{C}\) is the constant vector that we get when we do integration. The velocity vector could also be written \(\vec{v}(t)=\left[ \int{a_x(t)~dt} + C_x \right] \vhat{i} + \left[ \int{a_y(t)~dt} + C_y \right] \vhat{j} + \left[ \int{a_z(t)~dt} + C_z \right] \vhat{k} \) where \(\vec{C} = C_x\vhat{i} + C_y\vhat{j} + C_z\vhat{k}\).

Once we have the velocity vector (or if we are given the velocity vector), we can calculate the position vector. For the equations below, we assume the velocity vector is in the form \(\vec{v}(t) = v_x\vhat{i} + v_y\vhat{j} + v_z\vhat{k}\).

Velocity

\(\vec{v}(t) = v_x\vhat{i} + v_y\vhat{j} + v_z\vhat{k}\)

Position

\(\vec{r}(t) = \int{v_x~dt}\vhat{i} + \int{v_y~dt}\vhat{j} + \int{v_z~dt}\vhat{k} + \vec{K}\)

In each of the above equations we end up with general constants, in our case \(\vec{C}\) and \(\vec{K}\). You will probably run across problems that give you information that you can use to find the actual values of these constants. Most of the time the information is given in the form of initial conditions, i.e. values of velocity and/or position at time \(t=0\). However, values at any other time will also allow you to find the constants. To do this, you substitute the value for time into the final equation and evaluate. Some practice problems demonstrate how to do this.

Acceleration Vector Components

The acceleration vector \(\vec{a}(t)= x''(t)\vhat{i}+y''(t)\vhat{j}+z''(t)\vhat{k}\) can be expressed in terms of two other unit vectors, the unit tangent vector and the principal unit normal vector. After working some practice problems, you need to learn how to calculate the other two unit vectors before learning how to write the acceleration using them.

Practice

Find velocity and acceleration vectors when \(t = 0\) for \( \vec{r}(t) = e^{2t}\hat{i} + e^{-t}\hat{j} \)

Problem Statement

Find velocity and acceleration vectors when \(t = 0\) for \( \vec{r}(t) = e^{2t}\hat{i} + e^{-t}\hat{j} \)

Final Answer

velocity

\( \vec{v}(0) = \vec{r}'(0) = 2\hat{i} - \hat{j} \)

acceleration

\( \vec{a}(0) = \vec{r}''(0) = 4\hat{i} + \hat{j} \)

Problem Statement

Find velocity and acceleration vectors when \(t = 0\) for \( \vec{r}(t) = e^{2t}\hat{i} + e^{-t}\hat{j} \)

Solution

715 video

video by Krista King Math

Final Answer

velocity

\( \vec{v}(0) = \vec{r}'(0) = 2\hat{i} - \hat{j} \)

acceleration

\( \vec{a}(0) = \vec{r}''(0) = 4\hat{i} + \hat{j} \)

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Find velocity and acceleration at \( t=3/4 \) for \( \vec{r}(t) = 3\cos(2\pi t)\hat{i} + \) \( 3\sin(2\pi t)\hat{j} \)

Problem Statement

Find velocity and acceleration at \( t=3/4 \) for \( \vec{r}(t) = 3\cos(2\pi t)\hat{i} + \) \( 3\sin(2\pi t)\hat{j} \)

Final Answer

velocity

\( \vec{r}'(3/4) = 6\pi \hat{i} \)

acceleration

\( \vec{r}''(3/4) = 12\pi^2\hat{j} \)

Problem Statement

Find velocity and acceleration at \( t=3/4 \) for \( \vec{r}(t) = 3\cos(2\pi t)\hat{i} + \) \( 3\sin(2\pi t)\hat{j} \)

Solution

716 video

video by Krista King Math

Final Answer

velocity

\( \vec{r}'(3/4) = 6\pi \hat{i} \)

acceleration

\( \vec{r}''(3/4) = 12\pi^2\hat{j} \)

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As time \(t>0\) increases, a particle travels along the curve \( \mathcal{C} \) with position parameterized by \( \vec{r}(t) = 2(\cos t+t\sin t)\hat{i} + \) \( 2(\sin t-t\cos t)\hat{j} \). For each \(t>0\), compute the tangent (velocity) vector and the speed. What is the length of the curve which the particle travels over between \(t=0\) and \(t=4\pi\)?

Problem Statement

As time \(t>0\) increases, a particle travels along the curve \( \mathcal{C} \) with position parameterized by \( \vec{r}(t) = 2(\cos t+t\sin t)\hat{i} + \) \( 2(\sin t-t\cos t)\hat{j} \). For each \(t>0\), compute the tangent (velocity) vector and the speed. What is the length of the curve which the particle travels over between \(t=0\) and \(t=4\pi\)?

Hint

For the length of the curve, use the equation \( L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 } ~dt } \). See the vector functions arc length page for details.

Problem Statement

As time \(t>0\) increases, a particle travels along the curve \( \mathcal{C} \) with position parameterized by \( \vec{r}(t) = 2(\cos t+t\sin t)\hat{i} + \) \( 2(\sin t-t\cos t)\hat{j} \). For each \(t>0\), compute the tangent (velocity) vector and the speed. What is the length of the curve which the particle travels over between \(t=0\) and \(t=4\pi\)?

Hint

For the length of the curve, use the equation \( L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 } ~dt } \). See the vector functions arc length page for details.

Solution

In the middle of the video solution for this problem is a part that you were not asked to do in the problem statement, from about 8 minutes to about 10 and a half minutes in the video.

706 video

video by Dr Chris Tisdell

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Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i} + t^2\hat{j} + t^3\hat{k}\).

Problem Statement

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i} + t^2\hat{j} + t^3\hat{k}\).

Final Answer

velocity

\(\vec{v}(t) = \vec{r}'(t) = \hat{i} + 2t\hat{j} + 3t^2\hat{k}\)

acceleration

\(\vec{a}(t) = \vec{r}''(t) = 2\hat{j} + 6t\hat{k}\)

speed

\(\| \vec{v}(t) \| = \sqrt{1+4t^2+9t^4}\)

Problem Statement

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i} + t^2\hat{j} + t^3\hat{k}\).

Solution

712 video

video by Krista King Math

Final Answer

velocity

\(\vec{v}(t) = \vec{r}'(t) = \hat{i} + 2t\hat{j} + 3t^2\hat{k}\)

acceleration

\(\vec{a}(t) = \vec{r}''(t) = 2\hat{j} + 6t\hat{k}\)

speed

\(\| \vec{v}(t) \| = \sqrt{1+4t^2+9t^4}\)

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Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i}+3e^t\hat{j} + 4e^t\hat{k}\).

Problem Statement

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i}+3e^t\hat{j} + 4e^t\hat{k}\).

Final Answer

velocity

\( \vec{v}(t) = \vec{r}'(t) = \hat{i} + 3e^t\hat{j} + 4e^t\hat{k} \)

acceleration

\( \vec{a}(t) = \vec{r}''(t) = 3e^t\hat{j} + 4e^t\hat{k} \)

speed

\( \| \vec{v}(t) \| = \sqrt{1+25e^{2t}} \)

Problem Statement

Find velocity, acceleration and speed for the position function \(\vec{r}(t) = t\hat{i}+3e^t\hat{j} + 4e^t\hat{k}\).

Solution

713 video

video by Krista King Math

Final Answer

velocity

\( \vec{v}(t) = \vec{r}'(t) = \hat{i} + 3e^t\hat{j} + 4e^t\hat{k} \)

acceleration

\( \vec{a}(t) = \vec{r}''(t) = 3e^t\hat{j} + 4e^t\hat{k} \)

speed

\( \| \vec{v}(t) \| = \sqrt{1+25e^{2t}} \)

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Find velocity, acceleration and speed for the position function \( \vec{r}(t) = (3\cos t)\hat{i} + \) \( (3\sin t)\hat{j} - 4t\hat{k} \).

Problem Statement

Find velocity, acceleration and speed for the position function \( \vec{r}(t) = (3\cos t)\hat{i} + \) \( (3\sin t)\hat{j} - 4t\hat{k} \).

Final Answer

velocity

\( \vec{v}(t) = \vec{r}'(t) = (-3\sin t)\hat{i} + (3\cos t)\hat{j} - 4\hat{k} \)

acceleration

\( \vec{a}(t) = \vec{r}''(t) = (-3\cos t)\hat{i} - 3 \sin t \hat{j} \)

speed

\( \| \vec{v}(t) \| = 5 \)

Problem Statement

Find velocity, acceleration and speed for the position function \( \vec{r}(t) = (3\cos t)\hat{i} + \) \( (3\sin t)\hat{j} - 4t\hat{k} \).

Solution

714 video

video by Krista King Math

Final Answer

velocity

\( \vec{v}(t) = \vec{r}'(t) = (-3\sin t)\hat{i} + (3\cos t)\hat{j} - 4\hat{k} \)

acceleration

\( \vec{a}(t) = \vec{r}''(t) = (-3\cos t)\hat{i} - 3 \sin t \hat{j} \)

speed

\( \| \vec{v}(t) \| = 5 \)

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For the position function \(\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle\), find the velocity and acceleration vectors at the point \((\pi,2)\).

Problem Statement

For the position function \(\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle\), find the velocity and acceleration vectors at the point \((\pi,2)\).

Final Answer

\( \vec{v}(\pi) = \langle 2,0\rangle \); \( \vec{a}(\pi) = \langle 0,-1 \rangle \)

Problem Statement

For the position function \(\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle\), find the velocity and acceleration vectors at the point \((\pi,2)\).

Solution

The equations we need are \(\vec{v}(t)=\vec{r}'(t)\) and \(\vec{a}(t)=\vec{v}'(t)\). So the velocity vector is \(\vec{v}(t)=\langle 1-\cos t, \sin t\rangle\).
Now we need to find t. We are given the point \((\pi,2)\) on the position vector. So we have two equations we can use to find t,
\(t-\sin t=\pi\) and \(1-\cos t=2\). The second one is easier to solve for t, so we have
\(1-\cos t=2 \to \) \(\cos t=-1 \to \) \(t=\pi\)
Just to double-check, if we let \(t=\pi\) in the equation \(t-\sin t=\pi\), the answer checks.
So \(\vec{v}(\pi)=\langle 1-\cos \pi, \sin\pi\rangle\) giving us \(\vec{v}(\pi)=\langle 2,0 \rangle\).

To get the acceleration vector, we take the derivative of the velocity vector, i.e. \(\vec{a}(t)=\vec{v}'(t)\). So \(\vec{a}(t)=\langle \sin t, \cos t\rangle\) and at the point \((\pi,2)\) the acceleration vector is \(\vec{a}(\pi)=\langle 0,-1\rangle\).
Although the problem statement did not ask for a graph, here is a graph of the position function, the velocity vector and the acceleration vector.

Final Answer

\( \vec{v}(\pi) = \langle 2,0\rangle \); \( \vec{a}(\pi) = \langle 0,-1 \rangle \)

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Find the position vector function \(\vec{r}(t)\) for a particle with acceleration \(\vec{a}(t) = \langle 2t,2\sin(t),\cos(4t)\rangle\), initial velocity \(\vec{v}(0)=\langle 1,-3,2\rangle\) and initial position \(\vec{r}(0)=\langle 2,4,-1\rangle\).

Problem Statement

Find the position vector function \(\vec{r}(t)\) for a particle with acceleration \(\vec{a}(t) = \langle 2t,2\sin(t),\cos(4t)\rangle\), initial velocity \(\vec{v}(0)=\langle 1,-3,2\rangle\) and initial position \(\vec{r}(0)=\langle 2,4,-1\rangle\).

Final Answer

\( \vec{r}(t) = \langle (1/3)t^3+t+2, \) \( -2\sin(t)-t+4, \) \( (-1/16)\cos(4t)+2t-15/16\rangle \)

Problem Statement

Find the position vector function \(\vec{r}(t)\) for a particle with acceleration \(\vec{a}(t) = \langle 2t,2\sin(t),\cos(4t)\rangle\), initial velocity \(\vec{v}(0)=\langle 1,-3,2\rangle\) and initial position \(\vec{r}(0)=\langle 2,4,-1\rangle\).

Solution

2056 video

video by MIP4U

Final Answer

\( \vec{r}(t) = \langle (1/3)t^3+t+2, \) \( -2\sin(t)-t+4, \) \( (-1/16)\cos(4t)+2t-15/16\rangle \)

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Determine the velocity vector, speed and acceleration vector of an object when \(t=1\) given by the position vector \( \vec{r}(t) = t\vhat{i} + (-0.5t^2+4)\vhat{j} \).

Problem Statement

Determine the velocity vector, speed and acceleration vector of an object when \(t=1\) given by the position vector \( \vec{r}(t) = t\vhat{i} + (-0.5t^2+4)\vhat{j} \).

Final Answer

\( \vec{v}(1)=\langle 1,-1 \rangle\), speed=\(\sqrt{2} \), \( \vec{a}(1)=\langle 0,-1 \rangle \)

Problem Statement

Determine the velocity vector, speed and acceleration vector of an object when \(t=1\) given by the position vector \( \vec{r}(t) = t\vhat{i} + (-0.5t^2+4)\vhat{j} \).

Solution

2057 video

video by MIP4U

Final Answer

\( \vec{v}(1)=\langle 1,-1 \rangle\), speed=\(\sqrt{2} \), \( \vec{a}(1)=\langle 0,-1 \rangle \)

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Determine the velocity vector, speed and acceleration vector of an object when \(t=2\) for the position vector \( \vec{r}(t) = \cos(\pi t)\vhat{i} + \sin(\pi t)\vhat{j} + \) \( (t^2/2)\vhat{k} \)

Problem Statement

Determine the velocity vector, speed and acceleration vector of an object when \(t=2\) for the position vector \( \vec{r}(t) = \cos(\pi t)\vhat{i} + \sin(\pi t)\vhat{j} + \) \( (t^2/2)\vhat{k} \)

Final Answer

\( \vec{v}(2) = \langle 0,\pi,2 \rangle \), speed = \( \sqrt{\pi^2+4} \), \( \vec{a}(2) = \langle -\pi^2,0,1 \rangle \)

Problem Statement

Determine the velocity vector, speed and acceleration vector of an object when \(t=2\) for the position vector \( \vec{r}(t) = \cos(\pi t)\vhat{i} + \sin(\pi t)\vhat{j} + \) \( (t^2/2)\vhat{k} \)

Solution

2058 video

video by MIP4U

Final Answer

\( \vec{v}(2) = \langle 0,\pi,2 \rangle \), speed = \( \sqrt{\pi^2+4} \), \( \vec{a}(2) = \langle -\pi^2,0,1 \rangle \)

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Find the velocity and acceleration vectors and speed of the vector function \( \vec{r}(t) = \langle -5t,-3t^2,4t^4+3 \rangle \) at \(t=1\).

Problem Statement

Find the velocity and acceleration vectors and speed of the vector function \( \vec{r}(t) = \langle -5t,-3t^2,4t^4+3 \rangle \) at \(t=1\).

Final Answer

\( \vec{v}(t) = \langle -5,-6,16\rangle\), \( \vec{a}(t) = \langle 0,-6,48 \rangle \), speed = \( \sqrt{317} \)

Problem Statement

Find the velocity and acceleration vectors and speed of the vector function \( \vec{r}(t) = \langle -5t,-3t^2,4t^4+3 \rangle \) at \(t=1\).

Solution

2059 video

video by MIP4U

Final Answer

\( \vec{v}(t) = \langle -5,-6,16\rangle\), \( \vec{a}(t) = \langle 0,-6,48 \rangle \), speed = \( \sqrt{317} \)

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A car travels with a velocity vector given by \( \vec{v}(t) = \langle t^2,e^t+1 \rangle \), where t is measured in seconds and the vector components are measured in feet. If the initial position of the car is \( \vec{r}(0) = \langle 1,3 \rangle \), find the position of the car after one second.

Problem Statement

A car travels with a velocity vector given by \( \vec{v}(t) = \langle t^2,e^t+1 \rangle \), where t is measured in seconds and the vector components are measured in feet. If the initial position of the car is \( \vec{r}(0) = \langle 1,3 \rangle \), find the position of the car after one second.

Final Answer

\( \vec{r}(1) = \langle 4/3, e+3 \rangle \) feet

Problem Statement

A car travels with a velocity vector given by \( \vec{v}(t) = \langle t^2,e^t+1 \rangle \), where t is measured in seconds and the vector components are measured in feet. If the initial position of the car is \( \vec{r}(0) = \langle 1,3 \rangle \), find the position of the car after one second.

Solution

2060 video

video by PatrickJMT

Final Answer

\( \vec{r}(1) = \langle 4/3, e+3 \rangle \) feet

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A particle moves along a curve whose parametric equations are \( x(t) = e^{-t}\), \( y(t) = 2\cos(3t) \), \( z(t) = 2\sin(3t) \). (a) Determine the velocity and acceleration vectors.
(b) Find the magnitude of the velocity (speed) and acceleration at \(t=0\).

Problem Statement

A particle moves along a curve whose parametric equations are \( x(t) = e^{-t}\), \( y(t) = 2\cos(3t) \), \( z(t) = 2\sin(3t) \). (a) Determine the velocity and acceleration vectors.
(b) Find the magnitude of the velocity (speed) and acceleration at \(t=0\).

Final Answer

(a) \( \vec{v}(t) = -e^{-t}\vhat{i} - 6\sin(3t)\vhat{j} + 6\cos(3t)\vhat{k} \); \( \vec{a}(t) = e^{-t}\vhat{i} - 18\cos(3t)\vhat{j} - 18\sin(3t)\vhat{k} \)
(b) \( \|\vec{v}(0)\| = \sqrt{37} \); \( \|\vec{a}(0)\| = 5\sqrt{13} \)

Problem Statement

A particle moves along a curve whose parametric equations are \( x(t) = e^{-t}\), \( y(t) = 2\cos(3t) \), \( z(t) = 2\sin(3t) \). (a) Determine the velocity and acceleration vectors.
(b) Find the magnitude of the velocity (speed) and acceleration at \(t=0\).

Solution

2065 video

video by Dr Chris Tisdell

Final Answer

(a) \( \vec{v}(t) = -e^{-t}\vhat{i} - 6\sin(3t)\vhat{j} + 6\cos(3t)\vhat{k} \); \( \vec{a}(t) = e^{-t}\vhat{i} - 18\cos(3t)\vhat{j} - 18\sin(3t)\vhat{k} \)
(b) \( \|\vec{v}(0)\| = \sqrt{37} \); \( \|\vec{a}(0)\| = 5\sqrt{13} \)

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vector function - projectile motion 17calculus youtube playlist

Really UNDERSTAND Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Calculus: Early Transcendental Functions

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