Video: Derivatives of Vector-Valued Functions

In this video, we will learn how to determine the derivatives of vector-valued functions and find unit tangent vectors.

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Video Transcript

In this video, weโ€™re going to look at how we can use our understanding of calculus to find the derivative of a vector function in both two or three dimensions. Weโ€™ll also look at how to find the derivative of dot and cross products of two vector functions, find unit tangent vectors, and even consider higher-order derivatives of these functions.

The derivative ๐ซ prime of a vector function, ๐ซ, is to find much in the same way as for real-valued functions. Itโ€™s the limit as โ„Ž approaches zero of ๐ซ of ๐‘ก plus โ„Ž minus ๐ซ of ๐‘ก over โ„Ž, provided that limit exists. But what does this actually mean? We know that we can find the difference between two vectors by simply finding the difference of their component parts. So, if the vector ๐ซ is given by ๐‘“ of ๐‘ก, ๐‘” of ๐‘ก, โ„Ž of ๐‘ก, then ๐ซ of ๐‘ก plus โ„Ž minus ๐ซ of ๐‘ก is ๐‘“ of ๐‘ก plus โ„Ž minus ๐‘“ of ๐‘ก, ๐‘” of ๐‘ก plus โ„Ž minus ๐‘” of ๐‘ก, โ„Ž of ๐‘ก plus โ„Ž minus โ„Ž of ๐‘ก.

We also know that we can divide a vector by a constant by dividing each of the component parts by that constant. And of course, โ„Ž itself is a constant. And this is great news as it means we can find the derivative of a vector function by taking the derivative of each of its component parts.

In other words, if ๐ซ of ๐‘ก is some vector function ๐‘“ of ๐‘ก, ๐‘” of ๐‘ก, โ„Ž of ๐‘ก, where ๐‘“, ๐‘”, and โ„Ž are differentiable function, then ๐ซ prime of ๐‘ก is ๐‘“ prime of ๐‘ก, ๐‘” prime of ๐‘ก, โ„Ž prime of ๐‘ก. We can also see that if โ„Ž of ๐‘ก is equal to zero, โ„Ž prime of ๐‘ก is equal to zero. So, this method holds perfectly for vector functions in two dimensions, too. Letโ€™s have a look at an example.

Find the derivative of a vector-valued function ๐ซ of ๐‘ก equals one plus ๐‘ก cubed ๐ข plus five ๐‘ก squared plus one ๐ฃ plus ๐‘ก cubed plus two ๐ค.

Remember, we can find the derivative of a vector-valued function by taking the derivative of each component. That means weโ€™re individually going to differentiate, with respect to ๐‘ก, one plus ๐‘ก cubed, five ๐‘ก squared plus one, and ๐‘ก cubed plus two. We then recall that to differentiate polynomial terms, we multiply the entire term by the exponent and then reduce that exponent by one. The derivative of one is zero, and the derivative of ๐‘ก cubed is three ๐‘ก squared. So, differentiating our component for ๐ข and we get three ๐‘ก squared.

Next, weโ€™re going to differentiate the component for ๐ฃ. Thatโ€™s 10๐‘ก plus zero, which is simply 10๐‘ก. Finally, weโ€™re going to differentiate ๐‘ก cubed plus two with respect to ๐‘ก. Of course, the derivative of that constant is zero, so we obtain the derivative of ๐‘ก cubed plus two to be three ๐‘ก squared plus zero or just three ๐‘ก squared. So, the derivative ๐‘Ÿ prime of ๐‘ก is three ๐‘ก squared ๐ข plus 10๐‘ก ๐ฃ plus three ๐‘ก squared ๐ค.

So, really quite a straightforward process. And of course, the standard rules for differentiation apply. We can still use the chain rule, product rule, and quotient rule alongside the standard results for differentiating special functions such as trigonometric or logarithmic ones. Armed with this information, it follows that we should be able to find the equation of the tangent lines to these vector functions. Letโ€™s see what that might look like.

Calculate ๐‘“ prime of ๐‘  and find the vector form of the equation of the tangent line at ๐‘“ of zero for ๐‘“ of ๐‘  equals ๐‘  plus one, ๐‘  squared plus one, ๐‘  cubed plus one.

In this question, weโ€™ve been given a rule for ๐‘“ in terms of its ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-coordinates. We can alternatively think of this as a vector-valued function such that the vector ๐Ÿ of ๐‘  is equal to ๐‘  plus one, ๐‘  squared plus one, ๐‘  cubed plus one, where ๐‘  plus one is the ๐ข-component, ๐‘  squared plus one is the ๐ฃ-component, and ๐‘  cubed plus one is the ๐ค-component.

Remember then, to find the derivative of a vector-valued function, we simply take the derivative of each component. This means weโ€™re going to individually differentiate with respect to ๐‘ , ๐‘  plus one, ๐‘  squared plus one, and ๐‘  cubed plus one. That gives us one, two ๐‘ , and three ๐‘  squared, respectively. Taking this back into the rule for the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-coordinates, and we find ๐‘“ prime of ๐‘  to be equal to one, two s, three ๐‘  squared.

We do still need to find the vector form of the equation of the tangent line at ๐‘“ of zero. So, we recall that a line that passes through a point ๐‘ฅ nought, ๐‘ฆ nought, ๐‘ง nought with the direction vector ๐ฏ is given by the equation ๐ซ equals ๐ซ nought plus ๐‘ก times ๐ฏ. So, weโ€™ll need two things, a point that our line passes through and its direction of travel. We can find the latter by evaluating the derivative when the parameter value, ๐‘ , is equal to zero. So, letโ€™s substitute ๐‘  equals zero into our derivative. Thatโ€™s one, zero, zero or one ๐ข plus zero ๐ฃ plus zero ๐ค.

We can also find that the point our tangent line passes through by substituting ๐‘  equals zero into the original function. Thatโ€™s ๐‘“ of zero. When we do, we obtain ๐‘“ of zero to be zero plus one, zero plus one, zero plus one. Thatโ€™s one, one, one. And so, the vector form of the equation of the tangent line at ๐Ÿ zero, letโ€™s call that ๐ฟ, is one, one, one plus ๐‘ก times one, zero, zero.

There might even be occasions where weโ€™re required to find a unit tangent vector to a line. Letโ€™s think back to our previous example. We will begin in much the same way.

We begin by calculating the first derivative. Thatโ€™s ๐Ÿ prime of ๐‘ . Itโ€™s one, two ๐‘ , three ๐‘  squared. We then work out the magnitude of this derivative. Itโ€™s the square root of the sum of the squares of the component parts. So, itโ€™s the square root of one squared plus two ๐‘  squared plus three ๐‘  squared squared. Thatโ€™s the square root of one plus four ๐‘  squared plus nine ๐‘  to the fourth power. The unit tangent vector is the first derivative divided by the magnitude of that derivative. Here, thatโ€™s one over the square root of one plus four ๐‘  squared plus nine ๐‘  to the fourth power, two ๐‘  over that root, and three ๐‘  squared also over that root.

Okay, so far, so good. We know how to find the derivative of a vector-valued function, how to find the equation of the tangent line, and the unit tangent vector. But what about if weโ€™re dealing with a dot or a cross product of two vectors-valued functions? Well, the differentiation formula for real-valued functions have counterparts for vector-valued functions.

Suppose ๐ฎ and ๐ฏ are differentiable vector functions. We can say that the derivative of the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ prime and ๐ฏ plus the dot product of ๐ฎ and ๐ฏ prime. Similarly, the derivative of the cross product of ๐ฎ and ๐ฏ is equal to the cross product of ๐ฎ prime and ๐ฏ plus the cross product of ๐ฎ and ๐ฏ prime. Similarly, we can apply the chain rule to find the derivative of ๐ฎ of ๐‘“ of ๐‘ก. Itโ€™s ๐‘“ prime of ๐‘ก times ๐ฎ prime of ๐‘“ of ๐‘ก. Letโ€™s have a look at an example of one of these in action.

Calculate the derivative of the dot product of ๐ซ ๐‘ก and ๐ฌ ๐‘ก for the vector-valued functions ๐ซ of ๐‘ก equals sin ๐‘ก ๐ข plus cos ๐‘ก ๐ฃ and ๐ฌ of ๐‘ก equals cos ๐‘ก ๐ข plus sin ๐‘ก ๐ฃ. We can extend the product rule we know so well here and say that for differentiable vector-valued functions, ๐ฎ and ๐ฏ, the derivative of the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ prime and ๐ฏ plus the dot product of ๐ฎ and ๐ฏ prime. In our case, we can say that the derivative of the dot product of ๐ซ and ๐ฌ is equal to the dot product of ๐ซ prime and ๐ฌ plus the dot product of ๐ซ and ๐ฌ prime.

So, weโ€™ll begin by calculating ๐‘Ÿ prime of ๐‘ก and ๐‘  prime of ๐‘ก, remembering that to differentiate a vector-valued function, we differentiate each of its component parts. So, for ๐‘Ÿ prime of ๐‘ก, weโ€™re going to differentiate sin of ๐‘ก and cos of ๐‘ก. The derivative of sin ๐‘ก is cos ๐‘ก, and the derivative of cos ๐‘ก is negative sin ๐‘ก. So, we find ๐‘Ÿ prime of ๐‘ก to be equal to cos ๐‘ก ๐ข minus sin ๐‘ก ๐ฃ. To differentiate ๐‘  of ๐‘ก, weโ€™re going to differentiate cos of ๐‘ก and sin of ๐‘ก. That gives us a negative sin ๐‘ก ๐ข plus cos ๐‘ก ๐ฃ.

Next, weโ€™re going to find the dot product of ๐ซ prime of ๐‘ก and ๐ฌ of ๐‘ก. Itโ€™s cos ๐‘ก times cos ๐‘ก, which is cos squared ๐‘ก plus negative sin ๐‘ก multiplied by sin ๐‘ก. So, thatโ€™s negative sin squared ๐‘ก. We can rewrite this as simply cos two ๐‘ก. Letโ€™s find the dot product of ๐ซ of ๐‘ก and ๐ฌ prime of ๐‘ก. Itโ€™s sin ๐‘ก times negative sin ๐‘ก, which is negative sin squared ๐‘ก plus cos ๐‘ก times cos ๐‘ก. Thatโ€™s cos squared ๐‘ก again. We see that this is equal to negative sin squared ๐‘ก plus cos squared ๐‘ก, which, once again, is equal to cos two ๐‘ก.

The derivative of the dot product of ๐ซ and ๐ฌ is the sum of these. Itโ€™s cos two ๐‘ก plus cos two ๐‘ก, which is equal to two cos two ๐‘ก. Now, itโ€™s useful to know that weโ€™d actually get the same answer by computing the dot product and then taking the derivative. Letโ€™s check that. The dot product of ๐ซ and ๐ฌ is sin ๐‘ก times cos ๐‘ก plus cos ๐‘ก times sin ๐‘ก. Weโ€™ll find the derivative of sin ๐‘ก cos ๐‘ก plus cos ๐‘ก sin ๐‘ก.

Applying the product rule to sin ๐‘ก cos ๐‘ก and then cos ๐‘ก sin ๐‘ก, we get negative sin squared ๐‘ก plus cos squared ๐‘ก plus negative sin squared ๐‘ก plus cos squared ๐‘ก. And each of these is equal to cos of two ๐‘ก. So once again, we obtain the derivative of the dot product of ๐ซ and ๐ฌ to be two cos two ๐‘ก.

In our final example, weโ€™ll see how we can find the second derivative of a vector-valued function.

Find the second derivative of a vector-valued function, ๐ซ of ๐‘ก equals sin two ๐‘ก ๐ข minus cos ๐‘ก ๐ฃ plus ๐‘’ to the power of ๐‘ก ๐ค.

Remember, we find the derivative of a vector-valued function by taking the derivative of each component. This means we can find the first derivative, ๐ซ prime of ๐‘ก, by individually differentiating sin of two ๐‘ก, negative cos of ๐‘ก, and ๐‘’ of ๐‘ก with respect to ๐‘ก. The derivative of sin of two ๐‘ก is two cos two ๐‘ก. The derivative of negative cos ๐‘ก is sin ๐‘ก. And the derivative of ๐‘’ to the power of ๐‘ก is really nice. Itโ€™s just ๐‘’ to the power of ๐‘ก. So, weโ€™ve worked out ๐‘Ÿ prime of ๐‘ก, the first derivative.

We want to find the second derivative. Thatโ€™s the derivative of the derivative. And it follows that we can use the same process to achieve this, by differentiating ๐ซ prime of ๐‘ก. And this means we can individually, once again, differentiate each component. Weโ€™ll differentiate two cos two ๐‘ก, sin ๐‘ก, and ๐‘’ to the power of ๐‘ก with respect to ๐‘ก. The derivative of two cos two ๐‘ก is negative four cos two ๐‘ก. The derivative of sin ๐‘ก is cos ๐‘ก. And the derivative of ๐‘’ to the power of ๐‘ก is ๐‘’ to the power of ๐‘ก. So in vector form, the second derivative of our vector-valued function is negative four cos two ๐‘ก ๐ข plus cos ๐‘ก ๐ฃ plus ๐‘’ to the power of ๐‘ก ๐ค.

In this video, we saw that we can find the derivative of a vector-valued function by simply taking the derivative of each component. We saw that we could use the equation of a line given in vector form to find the vector equation of the tangent line to the function at a point, and that we can find the unit tangent vector by dividing the derivative function by the magnitude of the derivative. Finally, we saw that we can extend our usual rules for differentiating real-valued functions to vector-valued functions, as shown.

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