To find the vector in component form given the initial and terminal points, simply subtract the initial point from the terminal point.

Write the formula to find both the x and y-components of a vector.

Substitute the value of velocity and theta into the equations.

The vector is:  

In order to find the horizontal component, set up an equation involving cosine with 7 as the hypotenuse, since the side in the implied triangle that represents the horizontal component is adjacent to the 22-degree angle:

First, find the cosine of 22, then multiply by 7

To find the vertical component, set up an equation involving sine, since the side in the implied triangle that represents the vertical component is opposite the 22-degree angle:

First, find the sine of 22, then multiply by 7

We are almost done, but we need to make a small adjustment. The picture indicates that the vector points up and to the left, so the horizontal component, 6.49, should be negative:

First find the slope of the tangent to the line by taking the derivative.

Using the Exponential Rule we get the following,

 .

Then plug 1 into the equation as 1 is the point to find the slope at. 

.

One way of finding the slope at a given point is by finding the derivative. In this case, we can take the derivative of y with respect to x, and plug in the desired value for x.

Using the exponential rule we get the following derivative,

.

Plugging in x=2 from the point 2,3 gives us the final slope,

Thus our slope at the specific point is .

Note that in this case, using the y coordinate was not necessary.

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given.

The first derivative is

and for this function

and

So the slope is

To find the slope of the tangent line of the function at the given value, evaluate the first derivative for the given value.

The first derivative is

and for this function

and plugging in the specific x value we get,

So the slope is

.

Calculate the derivative of  by using the derivative rules.  The derivative function determines the slope at any point of the original function.

The derivative is: 

With the given point , .   Substitute this value to the derivative function to determine the slope at that point.

The slope of the tangent line that intersects point  is .

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point.  Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line.  Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point .  To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line.  Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for  will give us our slope-intercept form.

The equation of the tangent line at  depends on the derivative at that point and the function value.

The derivative at that point of  is 

 using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at  since , so that line is .