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In this video, weโre 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.
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Weโll 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.
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The derivative ๐ซ prime of a vector function, ๐ซ, is to find much in the same way as for real-valued functions.
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Itโs the limit as โ approaches zero of ๐ซ of ๐ก plus โ minus ๐ซ of ๐ก over โ, provided that limit exists.
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But what does this actually mean?
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We know that we can find the difference between two vectors by simply finding the difference of their component parts.
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So, if the vector ๐ซ is given by ๐ of ๐ก, ๐ of ๐ก, โ of ๐ก, then ๐ซ of ๐ก plus โ minus ๐ซ of ๐ก is ๐ of ๐ก plus โ minus ๐ of ๐ก, ๐ of ๐ก plus โ minus ๐ of ๐ก, โ of ๐ก plus โ minus โ of ๐ก.
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We also know that we can divide a vector by a constant by dividing each of the component parts by that constant.
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And of course, โ itself is a constant.
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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.
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In other words, if ๐ซ of ๐ก is some vector function ๐ of ๐ก, ๐ of ๐ก, โ of ๐ก, where ๐, ๐, and โ are differentiable function, then ๐ซ prime of ๐ก is ๐ prime of ๐ก, ๐ prime of ๐ก, โ prime of ๐ก.
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We can also see that if โ of ๐ก is equal to zero, โ prime of ๐ก is equal to zero.
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So, this method holds perfectly for vector functions in two dimensions, too.
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Letโs have a look at an example.
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Find the derivative of a vector-valued function ๐ซ of ๐ก equals one plus ๐ก cubed ๐ข plus five ๐ก squared plus one ๐ฃ plus ๐ก cubed plus two ๐ค.
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Remember, we can find the derivative of a vector-valued function by taking the derivative of each component.
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That means weโre individually going to differentiate, with respect to ๐ก, one plus ๐ก cubed, five ๐ก squared plus one, and ๐ก cubed plus two.
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We then recall that to differentiate polynomial terms, we multiply the entire term by the exponent and then reduce that exponent by one.
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The derivative of one is zero, and the derivative of ๐ก cubed is three ๐ก squared.
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So, differentiating our component for ๐ข and we get three ๐ก squared.
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Next, weโre going to differentiate the component for ๐ฃ.
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Thatโs 10๐ก plus zero, which is simply 10๐ก.
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Finally, weโre going to differentiate ๐ก cubed plus two with respect to ๐ก.
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Of course, the derivative of that constant is zero, so we obtain the derivative of ๐ก cubed plus two to be three ๐ก squared plus zero or just three ๐ก squared.
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So, the derivative ๐ prime of ๐ก is three ๐ก squared ๐ข plus 10๐ก ๐ฃ plus three ๐ก squared ๐ค.
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So, really quite a straightforward process.
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And of course, the standard rules for differentiation apply.
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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.
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Armed with this information, it follows that we should be able to find the equation of the tangent lines to these vector functions.
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Letโs see what that might look like.
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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.
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In this question, weโve been given a rule for ๐ in terms of its ๐ฅ-, ๐ฆ-, and ๐ง-coordinates.
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We can alternatively think of this as a vector-valued function such that the vector ๐ of ๐ is equal to ๐ plus one, ๐ squared plus one, ๐ cubed plus one, where ๐ plus one is the ๐ข-component, ๐ squared plus one is the ๐ฃ-component, and ๐ cubed plus one is the ๐ค-component.
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Remember then, to find the derivative of a vector-valued function, we simply take the derivative of each component.
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This means weโre going to individually differentiate with respect to ๐ , ๐ plus one, ๐ squared plus one, and ๐ cubed plus one.
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That gives us one, two ๐ , and three ๐ squared, respectively.
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Taking this back into the rule for the ๐ฅ-, ๐ฆ-, and ๐ง-coordinates, and we find ๐ prime of ๐ to be equal to one, two s, three ๐ squared.
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We do still need to find the vector form of the equation of the tangent line at ๐ of zero.
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So, we recall that a line that passes through a point ๐ฅ nought, ๐ฆ nought, ๐ง nought with the direction vector ๐ฏ is given by the equation ๐ซ equals ๐ซ nought plus ๐ก times ๐ฏ.
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So, weโll need two things, a point that our line passes through and its direction of travel.
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We can find the latter by evaluating the derivative when the parameter value, ๐ , is equal to zero.
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So, letโs substitute ๐ equals zero into our derivative.
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Thatโs one, zero, zero or one ๐ข plus zero ๐ฃ plus zero ๐ค.
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We can also find that the point our tangent line passes through by substituting ๐ equals zero into the original function.
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Thatโs ๐ of zero.
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When we do, we obtain ๐ of zero to be zero plus one, zero plus one, zero plus one.
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Thatโs one, one, one.
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And so, the vector form of the equation of the tangent line at ๐ zero, letโs call that ๐ฟ, is one, one, one plus ๐ก times one, zero, zero.
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There might even be occasions where weโre required to find a unit tangent vector to a line.
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Letโs think back to our previous example.
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We will begin in much the same way.
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We begin by calculating the first derivative.
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Thatโs ๐ prime of ๐ .
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Itโs one, two ๐ , three ๐ squared.
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We then work out the magnitude of this derivative.
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Itโs the square root of the sum of the squares of the component parts.
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So, itโs the square root of one squared plus two ๐ squared plus three ๐ squared squared.
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Thatโs the square root of one plus four ๐ squared plus nine ๐ to the fourth power.
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The unit tangent vector is the first derivative divided by the magnitude of that derivative.
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Here, thatโs 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.
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Okay, so far, so good.
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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.
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But what about if weโre dealing with a dot or a cross product of two vectors-valued functions?
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Well, the differentiation formula for real-valued functions have counterparts for vector-valued functions.
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Suppose ๐ฎ and ๐ฏ are differentiable vector functions.
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We can say that the derivative of the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ prime and ๐ฏ plus the dot product of ๐ฎ and ๐ฏ prime.
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Similarly, the derivative of the cross product of ๐ฎ and ๐ฏ is equal to the cross product of ๐ฎ prime and ๐ฏ plus the cross product of ๐ฎ and ๐ฏ prime.
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Similarly, we can apply the chain rule to find the derivative of ๐ฎ of ๐ of ๐ก.
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Itโs ๐ prime of ๐ก times ๐ฎ prime of ๐ of ๐ก.
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Letโs have a look at an example of one of these in action.
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Calculate the derivative of the dot product of ๐ซ ๐ก and ๐ฌ ๐ก for the vector-valued functions ๐ซ of ๐ก equals sin ๐ก ๐ข plus cos ๐ก ๐ฃ and ๐ฌ of ๐ก equals cos ๐ก ๐ข plus sin ๐ก ๐ฃ.
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We can extend the product rule we know so well here and say that for differentiable vector-valued functions, ๐ฎ and ๐ฏ, the derivative of the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ prime and ๐ฏ plus the dot product of ๐ฎ and ๐ฏ prime.
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In our case, we can say that the derivative of the dot product of ๐ซ and ๐ฌ is equal to the dot product of ๐ซ prime and ๐ฌ plus the dot product of ๐ซ and ๐ฌ prime.
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So, weโll begin by calculating ๐ prime of ๐ก and ๐ prime of ๐ก, remembering that to differentiate a vector-valued function, we differentiate each of its component parts.
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So, for ๐ prime of ๐ก, weโre going to differentiate sin of ๐ก and cos of ๐ก.
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The derivative of sin ๐ก is cos ๐ก, and the derivative of cos ๐ก is negative sin ๐ก.
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So, we find ๐ prime of ๐ก to be equal to cos ๐ก ๐ข minus sin ๐ก ๐ฃ.
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To differentiate ๐ of ๐ก, weโre going to differentiate cos of ๐ก and sin of ๐ก.
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That gives us a negative sin ๐ก ๐ข plus cos ๐ก ๐ฃ.
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Next, weโre going to find the dot product of ๐ซ prime of ๐ก and ๐ฌ of ๐ก.
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Itโs cos ๐ก times cos ๐ก, which is cos squared ๐ก plus negative sin ๐ก multiplied by sin ๐ก.
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So, thatโs negative sin squared ๐ก.
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We can rewrite this as simply cos two ๐ก.
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Letโs find the dot product of ๐ซ of ๐ก and ๐ฌ prime of ๐ก.
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Itโs sin ๐ก times negative sin ๐ก, which is negative sin squared ๐ก plus cos ๐ก times cos ๐ก.
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Thatโs cos squared ๐ก again.
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We see that this is equal to negative sin squared ๐ก plus cos squared ๐ก, which, once again, is equal to cos two ๐ก.
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The derivative of the dot product of ๐ซ and ๐ฌ is the sum of these.
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Itโs cos two ๐ก plus cos two ๐ก, which is equal to two cos two ๐ก.
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Now, itโs useful to know that weโd actually get the same answer by computing the dot product and then taking the derivative.
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Letโs check that.
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The dot product of ๐ซ and ๐ฌ is sin ๐ก times cos ๐ก plus cos ๐ก times sin ๐ก.
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Weโll find the derivative of sin ๐ก cos ๐ก plus cos ๐ก sin ๐ก.
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Applying the product rule to sin ๐ก cos ๐ก and then cos ๐ก sin ๐ก, we get negative sin squared ๐ก plus cos squared ๐ก plus negative sin squared ๐ก plus cos squared ๐ก.
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And each of these is equal to cos of two ๐ก.
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So once again, we obtain the derivative of the dot product of ๐ซ and ๐ฌ to be two cos two ๐ก.
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In our final example, weโll see how we can find the second derivative of a vector-valued function.
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Find the second derivative of a vector-valued function, ๐ซ of ๐ก equals sin two ๐ก ๐ข minus cos ๐ก ๐ฃ plus ๐ to the power of ๐ก ๐ค.
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Remember, we find the derivative of a vector-valued function by taking the derivative of each component.
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This means we can find the first derivative, ๐ซ prime of ๐ก, by individually differentiating sin of two ๐ก, negative cos of ๐ก, and ๐ of ๐ก with respect to ๐ก.
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The derivative of sin of two ๐ก is two cos two ๐ก.
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The derivative of negative cos ๐ก is sin ๐ก.
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And the derivative of ๐ to the power of ๐ก is really nice.
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Itโs just ๐ to the power of ๐ก.
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So, weโve worked out ๐ prime of ๐ก, the first derivative.
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We want to find the second derivative.
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Thatโs the derivative of the derivative.
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And it follows that we can use the same process to achieve this, by differentiating ๐ซ prime of ๐ก.
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And this means we can individually, once again, differentiate each component.
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Weโll differentiate two cos two ๐ก, sin ๐ก, and ๐ to the power of ๐ก with respect to ๐ก.
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The derivative of two cos two ๐ก is negative four cos two ๐ก.
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The derivative of sin ๐ก is cos ๐ก.
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And the derivative of ๐ to the power of ๐ก is ๐ to the power of ๐ก.
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So in vector form, the second derivative of our vector-valued function is negative four cos two ๐ก ๐ข plus cos ๐ก ๐ฃ plus ๐ to the power of ๐ก ๐ค.
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In this video, we saw that we can find the derivative of a vector-valued function by simply taking the derivative of each component.
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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.
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Finally, we saw that we can extend our usual rules for differentiating real-valued functions to vector-valued functions, as shown.