Question Video: Finding the Tangent to a Vector-Valued Function

Calculate πβ²(π ), and find the vector form of the equation of the tangent line πΏ at π(0) for π(π ) = (cos 2π , sin 2π , π ).

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Video Transcript

Calculate π prime of π , and find the vector form of the equation of the tangent line πΏ at π of zero for π of π  equals cos of two π , sin of two π , π .

We can think of this as a vector-valued function. And the first thing weβre asked to do is to find π prime of π . Thatβs the first derivative of this function. Now, this is fairly straightforward. All we need to do to find the derivative of a vector function is take the derivative of each of our components. Thatβs the derivative of cos of two π , the derivative of sin of two π , and the derivative of π , all with respect to π .

So, weβll begin by quoting the general result for the derivative of cos of ππ₯. Itβs negative π sin of ππ₯. And this means that the derivative of cos of two π , with respect to π , is a negative two sin of two π . Similarly, the general result for the derivative of sin ππ₯ is π cos ππ₯. So, d by dπ  of sin two π  is two cos of two π .

We also know that we can differentiate a polynomial term by multiplying it by its exponent and then reducing the exponent by one. Well, the exponent for π  is technically one. So, itβs one times π  to the power of zero. But π  to the power of zero is one. So, the first derivative of π  with respect to π  is just one. And this means, π prime of π  is equal to negative two sin two π , two cos two π , one.

Our next job is to find the vector form of the equation of the tangent line πΏ at π of zero. In other words, when our parameter π  is equal to zero. Well, letβs recall the vector equation for a straight line. A line that passes through a point π which has position vector π nought and has a direction vector π£ is given by π equals π nought plus π‘ times π£. So, weβre going to need to do two things. Weβre going to need to work out a point that our tangent line passes through and its direction of travel.

Well, we can find the direction of travel by evaluating the derivative when our parameter π  is equal to zero. So, thatβs π prime of zero. π prime of zero is negative two sin of two times zero, two cos of two times zero, one. Thatβs zero, two, one. We could alternatively write this as π£ equals two π plus π.

But how do we find the point that our tangent line passes through? Well, thatβs quite straightforward too. All we need to do is evaluate our function when π , our parameter, is equal to zero. Thatβs cos of two times zero, sin of two times zero, zero. Which is one, zero, zero. And so, we see that the vector form of the equation of our tangent line is one, zero, zero plus π‘ times zero, two, one.