Video: Finding the Equation of a Tangent Line to a Parametrically Defined Curve

The position of a particle is given by the parametric equations π‘₯(𝑑) = 5√(2𝑑 βˆ’ 3) and 𝑦(𝑑) = 12βˆ›(𝑑² + 4). Find the slope of the tangent line to the path of the particle at the time 𝑑 = 2 seconds.

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Video Transcript

The position of a particle is given by the parametric equations π‘₯ of 𝑑 equals five square root of two 𝑑 minus three and 𝑦 of 𝑑 equals 12 cube root of 𝑑 squared plus four. Find the slope of the tangent line to the path of the particle at the time 𝑑 equals two seconds.

When looking for the slope of a tangent line of a curve, we’re interested in finding the derivative. In this question though, we’ve been given a pair of parametric equations. These are equations for π‘₯ and 𝑦 in terms of a third parameter 𝑑. And so we recall that as long as dπ‘₯ by d𝑑 is not equal to zero, d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑑 divided by dπ‘₯ by d𝑑. Now, let’s rewrite our equations for π‘₯ and 𝑦. We can say that π‘₯ is equal to five times two 𝑑 minus three to the power of one-half. And 𝑦 is equal to 12 times 𝑑 squared plus four to the power of one-third.

π‘₯ and 𝑦 are composite functions. That is, they are a function of another function. And so we’re going to use the chain rule to find their derivatives. Now, the chain rule says that if 𝑦 is some function in 𝑒 and 𝑒 is some function in π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. Now, of course, we’re actually initially looking to find dπ‘₯ by d𝑑. So let’s change our chain rule slightly. This time, we can say that dπ‘₯ by d𝑑 is equal to dπ‘₯ by d𝑒 times d𝑒 by d𝑑. And we’re going to let 𝑒 be equal to the inner part of our composite function. It’s two 𝑑 minus three so that π‘₯ is equal to five 𝑒 to the power of one-half. We can differentiate our function 𝑒 with respect to 𝑑, and we get two.

Then, to differentiate π‘₯ with respect to 𝑒, we multiply our entire term by the exponent and reduce that exponent by one. So we get a half times five 𝑒 to the power of negative one-half, which is five over two times 𝑒 to the power of negative one-half. Now, dπ‘₯ by d𝑑 is the product of these. So it’s two times five over two 𝑒 to the power of negative one-half. And we can see that the twos cancel. By recognizing that 𝑒 to the power of negative one-half is one over the square root of 𝑒, we write this further as five over the square root of 𝑒. But, of course, we were looking to differentiate π‘₯ with respect to 𝑑. So we’re going to go back to our substitution and replace 𝑒 with two 𝑑 minus three. So dπ‘₯ by d𝑑 is five over the square root of two 𝑑 minus three.

We rewrite our chain rule once again. And we’re going to find d𝑦 by d𝑑. This time, we let 𝑒 be equal to 𝑑 squared plus four. Once again, that is the inner part of our composite function. And 𝑦 is therefore 12𝑒 to the power of one-third. d𝑒 by d𝑑 is fairly straightforward. It’s two 𝑑. Whereas d𝑦 by d𝑒 is a third times 12𝑒 to the power of negative two-thirds, which is four 𝑒 to the power of negative two-thirds. Now, again, d𝑦 by d𝑑 is the product of these. It’s two 𝑑 times four 𝑒 to the power of negative two-thirds. 𝑒 to the power of negative two-thirds is one over 𝑒 to the power of two-thirds. So we get eight 𝑑 over 𝑒 to the power of two-thirds. But once again, we want to replace 𝑒 with our original substitution. And we therefore find d𝑦 by d𝑑 to be equal to eight 𝑑 over 𝑑 squared plus four to the power of two-thirds.

d𝑦 by dπ‘₯ is the quotient of these. It’s d𝑦 by d𝑑 divided by dπ‘₯ by d𝑑. And we could look to simplify this a little. But remember, we’re trying to find the slope of the tangent line to the path of the particle when 𝑑 is equal to two. And so we substitute 𝑑 equals two into our expression for the derivative. And we get eight times two over two squared plus four to the power of two-thirds divided by five over the square root of two times two minus three. This simplifies to 16 over eight to the power of two-thirds divided by five divided by the square root of one.

But, of course, eight to the power of two-thirds is the cube root of eight squared. Now, the cube root of eight is two. So we get two squared, which is equal to four. The square root of one is one, so we get 16 over four divided by five. But 16 divided by four is four. So the derivative at 𝑑 equals two is four divided by five, which is, of course, four-fifths.

And so, given this pair of parametric equations, the slope of the tangent line to the path of the particle at the time 𝑑 equals two seconds is four-fifths.

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