Question Video: Find the Derivative of the Dot Product of Two Vectors | Nagwa Question Video: Find the Derivative of the Dot Product of Two Vectors | Nagwa

Question Video: Find the Derivative of the Dot Product of Two Vectors Mathematics

Calculate (d/d𝑑) (𝐫(𝑑) β‹… 𝐬(𝑑)) for the vector-valued functions 𝐫(𝑑) = sin 𝑑𝐒 + cos 𝑑𝐣 and 𝐬(𝑑) = cos 𝑑𝐒 + sin 𝑑𝐣.

02:54

Video Transcript

Calculate the derivative of 𝐫 of 𝑑 dot 𝐬 of 𝑑 with respect to 𝑑 for the vector-valued functions 𝐫 of 𝑑 is equal to the sin of 𝑑 times 𝐒 plus the cos of 𝑑 times 𝐣 and 𝐬 of 𝑑 is equal to the cos of 𝑑 times 𝐒 plus the sin of 𝑑 times 𝐣.

The question gives us two vector-valued functions, 𝐫 of 𝑑 and 𝐬 of 𝑑. It wants us to find the derivative with respect to 𝑑 of 𝐫 of 𝑑 dot 𝐬 of 𝑑. Remember, since these are vector-valued functions, the dot does not mean multiply; it means the dot product. Before we move any further with this question, it’s also worth pointing out some notations about vectors.

There’s a lot of different ways you can see vectors written. For example, you can see them written with bold notation, with underlines, or with full arrows. In this video, we’re going to use half-arrow notation to represent vectors. We’re also going to use another notation, which is the hat notation. This represents vectors with unit length. Let’s now recall what we mean by the dot product of two vectors.

We recall π‘Žπ’ plus 𝑏𝐣 dot-producted with 𝑐𝐒 plus 𝑑𝐣 will give us π‘Žπ‘ plus 𝑏𝑑. In other words, we multiply componentwise and then add the results together. So, let’s use this to find an expression for 𝐫 of 𝑑 dot 𝐬 of 𝑑. First, we need to multiply the first components in both expressions together. That’s the sin of 𝑑 times the cos of 𝑑.

Next, we need to add to this the product of the second components of each of our expressions. That’s the cos of 𝑑 times the sin of 𝑑. So, we get that 𝐫 of 𝑑 dot-producted with 𝐬 of 𝑑 is equal to sin of 𝑑 cos of 𝑑 plus cos of 𝑑 sin of 𝑑. Of course, since multiplication is commutative, we can simplify this by rearranging our second term and then adding these together. And doing this, we get two times the sin of 𝑑 times the cos of 𝑑.

Remember, we’re not done though. The question wants us to differentiate this with respect to 𝑑. And we could do this directly by using the product rule. However, there’s another way. Recall the double-angle formula for sine tells us the sin of two 𝑑 is equivalent to two times the sin of 𝑑 cos of 𝑑. In other words, we can write 𝐫 of 𝑑 dot 𝐬 of 𝑑 as the sin of two 𝑑. And this expression is much easier to differentiate.

So, we now have the derivative of 𝐫 of 𝑑 dot 𝐬 of 𝑑 with respect to 𝑑 is equal to the derivative of the sin of two 𝑑 with respect to 𝑑. And of course, we know for any constant π‘Ž, the derivative of the sin of π‘Žπ‘‘ is equal to π‘Ž times the cos of π‘Žπ‘‘. And in our case, the value of the constant π‘Ž is equal to two. So, we get two times the cos of two 𝑑.

Therefore, given the vector-valued functions 𝐫 of 𝑑 is equal to the sin of 𝑑𝐒 plus the cos of 𝑑𝐣 and 𝐬 of 𝑑 is equal to the cos of 𝑑𝐒 plus the sin of 𝑑𝐣. We were able to show the derivative of 𝐫 of 𝑑 dot 𝐬 of 𝑑 with respect to 𝑑 is equal to two times the cos of two 𝑑.

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