Video: MATH-DIFF-INT-2018-S1-Q05

Let π‘₯ = 2𝑑³ βˆ’ 15𝑑² + 36𝑑 βˆ’ 1 and 𝑦 = 𝑑² βˆ’ 8𝑑 + 11. For which values of 𝑑 does this curve have a vertical tangent?

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Video Transcript

Let π‘₯ equal two 𝑑 cubed two minus 15𝑑 squared plus 36𝑑 minus one and 𝑦 equal 𝑑 squared minus eight 𝑑 plus 11. For which values of 𝑑 does this curve have a vertical tangent?

Let’s start by working out what the question is asking us to find, mathematically, when it asks us to find the values of 𝑑 where this curve has a vertical tangent. Now, if the question had’ve asked us to find a horizontal tangent, what we’d have been looking for is points on the curve where the tangent to the curve is horizontal. Now, these will be the points where the gradient of the line is zero. And so, we can say that at these points, d𝑦 by dπ‘₯ will be equal to zero. Or, the change in 𝑦 with respect to π‘₯ is zero.

Now, the question asks us to find the values of 𝑑 for which the curve has a vertical tangent and not a horizontal tangent. So what this means is the points on the line where the tangent to that point is vertical. And so, we can say that at these points, the change in π‘₯ with respect to 𝑦 is zero. In mathematical terms, this means that dπ‘₯ by d𝑦 is equal to zero, since dπ‘₯ by d𝑦 represents the change in π‘₯ with respect to 𝑦. In order to find the values of 𝑑 for which the curve has vertical tangent, we can simply find the values of 𝑑 so that dπ‘₯ by d𝑦 is equal to zero.

Now, the equation for our curve is in parametric equations. And in order to find d𝑦 by dπ‘₯ or dπ‘₯ by d𝑦 in terms of parametrics, we will use the following equation. We have that d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑑 over dπ‘₯ by d𝑑. From this, we obtain that dπ‘₯ by d𝑦 is equal to dπ‘₯ by d𝑑 over d𝑦 by d𝑑. In our case, we have that dπ‘₯ by d𝑦 is equal to zero. And since dπ‘₯ by d𝑦 is also equal to dπ‘₯ by d𝑑 over d𝑦 by d𝑑, we obtain that dπ‘₯ by d𝑑 over d𝑦 by d𝑑 is equal to zero.

Next, we will use the fact that if any fraction is equal to zero, then this means that the numerator must be equal to zero. And so, therefore, this tells us that dπ‘₯ by d𝑑 is equal to zero. The next step is to find dπ‘₯ by d𝑑. We’re given in the question that π‘₯ is equal to two 𝑑 cubed minus 15𝑑 squared plus 36𝑑 minus one. And this is simply a polynomial. When differentiating polynomial terms, we simply multiply by the power and decrease the power by one.

For the first term here, we have two 𝑑 cubed. So when we differentiate this with respect to 𝑑, we multiply by the power, so that’s multiplying by three. And a two times three gives us a six. Then, we decrease the power by one, so 𝑑 cubed goes to 𝑑 squared, leaving us with six 𝑑 squared. For the next term, we have negative 15𝑑 squared. So we multiply it by the power to give us negative 30. Then we decrease the power by one. So we end up with 𝑑 to the power of one, which is simply 𝑑. And so, this term becomes negative 30𝑑. Then, the next term is 36𝑑. And we multiply it by the power of 𝑑, which is just one, to give us 36 and decrease the power of 𝑑 by one. So that becomes 𝑑 to the power of zero or just one.

Then, the last term here is negative one, which is simply a constant. And when we differentiate a constant by anything, it always becomes zero. And so, this term differentiates to zero. However, adding zero does absolutely nothing here. So we can just ignore it. Now, we have found dπ‘₯ by d𝑑. We can set it equal to zero to give us that six 𝑑 squared minus 30𝑑 plus 36 is equal to zero. We spot that each term here is a multiple of six. So we can divide the whole equation by six, which gives us 𝑑 squared minus five 𝑑 plus six is equal to zero.

Factorising this gives us 𝑑 minus two times 𝑑 minus three is equal to zero. So we obtain that 𝑑 minus two equals zero, which gives us that 𝑑 is equal to two. Or, 𝑑 minus three is equal to zero, which gives us that 𝑑 is equal to three. And so, we have found the values of 𝑑 for which dπ‘₯ by d𝑑 is equal to zero. And, therefore, these values of 𝑑 also give dπ‘₯ by d𝑦 is equal to zero. And so, these must be the values of 𝑑 for which the curve has a vertical tangent, giving us a solution of 𝑑 is equal to two or 𝑑 is equal to three.

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