Video: Differentiating Exponential Functions Using the Chain Rule

A particle has a position defined by the equations π‘₯ = 𝑑³ βˆ’ 5𝑑 and 𝑦 = 3 βˆ’ 2𝑑². Find the velocity vector of the particle at 𝑑 = 2.

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Video Transcript

A particle has a position defined by the equations π‘₯ equals 𝑑 cubed minus five 𝑑 and 𝑦 equals three minus two 𝑑 squared. Find the velocity vector of the particle at 𝑑 equals two.

We’re told that the position of the particle is defined by a pair of parametric equations. In real terms, this means that, given a value of time 𝑑, we obtain a coordinate pair π‘₯𝑦 for the position of the particle at that time. We could say that, in vector terms, the position of the particle at time 𝑑 is 𝑠 of 𝑑 equals 𝑑 cubed minus five 𝑑 𝑖 plus three minus two 𝑑 squared 𝑗. This isn’t though quite what we’re looking for. We want to find the velocity vector of our particle when 𝑑 is equal to two. So we recall that, given a function for displacement which of course is essentially the difference in the object’s position from one time to another, we find a function for velocity by differentiating with respect to 𝑑.

This means that we can achieve a velocity vector for our particle at time 𝑑 by differentiating each component for the position vector with respect to 𝑑. That’s the derivative of 𝑑 cubed minus five 𝑑. And then we differentiate the vertical component as well, the derivative of three minus two 𝑑 squared. We know that to differentiate a polynomial term, we multiply the entire term by the experiment and then reduce the exposure by one. So 𝑑 cubed differentiates to three 𝑑 squared and negative five 𝑑 differentiates to negative five.

Similarly, the derivative of three minus two 𝑑 squared is negative four 𝑑. Remember, the derivative of any constant is simply zero. Our vector function for velocity is three 𝑑 squared minus five 𝑖 plus negative four 𝑑 𝑗. Remember, we want a velocity vector at 𝑑 is equal to two. So we’re going to substitute 𝑑 equals two into our vector. That’s three times two squared minus five 𝑖 plus negative four times two 𝑗, which simplifies to seven 𝑖 minus eight 𝑗.

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