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 • Higher Education

Calculate (d/dπ‘) (π«(π‘) β π¬(π‘)) for the vector-valued functions π«(π‘) = sin π‘π’ + cos π‘π£ and π¬(π‘) = cos π‘π’ + sin π‘π£.

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Video Transcript

Calculate the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ for the vector-valued functions π« of π‘ is equal to the sin of π‘ times π’ plus the cos of π‘ times π£ and π¬ of π‘ is equal to the cos of π‘ times π’ plus the sin of π‘ times π£.

The question gives us two vector-valued functions, π« of π‘ and π¬ of π‘. It wants us to find the derivative with respect to π‘ of π« of π‘ dot π¬ of π‘. 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, itβs also worth pointing out some notations about vectors.

Thereβ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, weβre going to use half-arrow notation to represent vectors. Weβre also going to use another notation, which is the hat notation. This represents vectors with unit length. Letβs now recall what we mean by the dot product of two vectors.

We recall ππ’ plus ππ£ dot-producted with ππ’ plus ππ£ will give us ππ plus ππ. In other words, we multiply componentwise and then add the results together. So, letβs use this to find an expression for π« of π‘ dot π¬ of π‘. First, we need to multiply the first components in both expressions together. Thatβs the sin of π‘ times the cos of π‘.

Next, we need to add to this the product of the second components of each of our expressions. Thatβs the cos of π‘ times the sin of π‘. So, we get that π« of π‘ dot-producted with π¬ of π‘ is equal to sin of π‘ cos of π‘ plus cos of π‘ sin of π‘. 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 π‘ times the cos of π‘.

Remember, weβre not done though. The question wants us to differentiate this with respect to π‘. And we could do this directly by using the product rule. However, thereβs another way. Recall the double-angle formula for sine tells us the sin of two π‘ is equivalent to two times the sin of π‘ cos of π‘. In other words, we can write π« of π‘ dot π¬ of π‘ as the sin of two π‘. And this expression is much easier to differentiate.

So, we now have the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ is equal to the derivative of the sin of two π‘ with respect to π‘. And of course, we know for any constant π, the derivative of the sin of ππ‘ is equal to π times the cos of ππ‘. And in our case, the value of the constant π is equal to two. So, we get two times the cos of two π‘.

Therefore, given the vector-valued functions π« of π‘ is equal to the sin of π‘π’ plus the cos of π‘π£ and π¬ of π‘ is equal to the cos of π‘π’ plus the sin of π‘π£. We were able to show the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ is equal to two times the cos of two π‘.

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