All Calculus 2 Resources
Example Questions
Example Question #1 : Derivatives Of Vectors
Given the following vector:
Find the second derivative of .
In order to obtain the second derivative, we will have to differentiate each component twice. We know how to differentiate the following:
.
We also have :
and this gives:
For the constant component , we know that it is derivative is zero.
This gives us the solution that we are looking for:
Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions
Given that . We define its gradient as :
Let be given by:
What is the gradient of ?
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 :
To see this, fix all other variables and assume that you have only as the only variable.
Now we apply the given defintion , i.e,
with :
this gives us the solution .
Example Question #7 : Derivatives Of Parametric, Polar, And Vector Functions
Let .
We define the gradient of as:
Let .
Find the vector gradient.
We note first that :
Using the Chain Rule where is the only variable here.
Using the Chain Rule where is the only variable here.
Continuing in this fashion we have:
Again using the Chain Rule and assuming that is the variable and all the others are constant.
Now applying the given definition of the gradient we have the required result.
Example Question #441 : Ap Calculus Bc
Let
What is the derivative of ?
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.
is the derivative of the first component.
of the second component.
is the derivative of the last component . we obtain then:
Example Question #1 : Derivatives Of Vectors
Let be given by:
Find the derivative of .
To find the derivative of u, we will have to find the derivative of every component.
Using the Chain Rule, we note that :
Continuing in the same fashion we have :
Ordering these components pairwise , we have then:
this is what we needed to show.
Example Question #2 : Derivatives Of Vectors
Let be given by:
with .
What is the value of ?
To obtain the desired result, we will have to differentiate the expression of u with respect to each variable and add the derivatives.
We have
we can also write the above expression as :
Now summing up we have:
But we are given that :
This gives:
Example Question #3 : Derivatives Of Vectors
Let and
Let . What is the derivative of ?
One way to obtain the solution is that we first add the two functions to obtain and we differentiate with each component to get the result.
Adding the two functions we obtain:
Now we will differentialte each component to get the derivative of .
We have :
Example Question #4 : Derivatives Of Vectors
We let with .
What is the derivative of order of ?
To obtain the derivative we will need to compute the derivative of each component. Note that we have a polynomial in this case.
We are also given that and with m,n and s positive integers.
We know that if given , then
so for k=n, we have
and
for m< n .
In the same manner we have :
This shows that the derivative of order n is given by:
Example Question #991 : Calculus Ii
Assuming that is the vector position of a moving vehicle. Can the velocity be zero?
Yes at
Yes at
Never
Yes at
Never
We first have to determine the expression of the velocity and see what happens as we move toward infinity.
Let us first compute the derivative. To do that, we do it componentwize.
We have,
Therefore the expression of the velocity .
Since we have , , for all t.
The velocity can never be zero.
Example Question #1 : Derivatives Of Vectors
Let
.
What are the values of for which is defined?
To have defined, we need to have all the components defined on the same interval, for
is defined for all t.
is defined if
This gives