Video: AP Calculus AB Exam 1 β€’ Section II β€’ Part B β€’ Question 4

The equation of a curve is given as 2𝑦² βˆ’ π‘₯Β² = βˆ’16𝑦 βˆ’ 36. i. Find d𝑦/dπ‘₯ in terms of π‘₯ and 𝑦. ii. Write an equation for the line tangent to the curve at the point (6, 0). iii. Write an equation of each vertical tangent to the curve. iv. Evaluate d2𝑦/dπ‘₯Β² at the point (3√6, 1).

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Video Transcript

The equation of a curve is given as two 𝑦 squared minus π‘₯ squared is equal to negative 16𝑦 minus 36. Part one) Find d𝑦 by dπ‘₯ in terms of π‘₯ and 𝑦. Part two) Write an equation for the line tangent to the curve at the point six, zero. Part three) Write an equation of each vertical tangent to the curve. Part four) Evaluate d two 𝑦 by dπ‘₯ squared at the point three root six, one.

For part one of this question, we’re asked to find d𝑦 by dπ‘₯ in terms of π‘₯ and 𝑦. Our equation is two 𝑦 squared minus π‘₯ squared is equal to negative 16𝑦 minus 36. We can find d𝑦 by dπ‘₯ by using implicit differentiation. This means we simply differentiate the whole equation with respect to π‘₯. We can do this term by term for each term in the equation. We’ll start with the term two 𝑦 squared. We need to find the differential of two 𝑦 squared with respect to π‘₯.

In order to do this, we require the chain rule. We can recall that the chain rule tells us that d𝑓 by dπ‘₯ is equal to d𝑓 by d𝑦 multiplied by d𝑦 by dπ‘₯. If we let two 𝑦 squared here be equal to 𝑓, then we can say that our differential here is equal to d by d𝑦 of two 𝑦 squared multiplied by d𝑦 by dπ‘₯. We can find d by d𝑦 of two 𝑦 squared by using the power rule for differentiation. We multiply by the power and decrease the power by one. We obtain that the differential of our first term is four 𝑦 multiplied by d𝑦 by dπ‘₯.

The second term in our equation is negative π‘₯ squared. And we can differentiate this with respect to π‘₯ by using the power rule for differentiation. We obtain negative two π‘₯. Now moving to the right-hand side of the equation, our next term is negative 16𝑦. When we differentiate 𝑦 with respect to π‘₯, we simply get d𝑦 by dπ‘₯. And we mustn’t forget to multiply it by our constant. This gives us negative 16 d𝑦 by dπ‘₯. Our final term negative 36 is a constant. And when we differentiate any constant, we’ll get zero. Therefore, we have finished differentiating our equation with respect to π‘₯.

Now, we simply need to make d𝑦 by dπ‘₯ the subject. First, we divide the whole equation by two. Then, we move all the terms containing d𝑦 by dπ‘₯ to one side of the equation. Then, we factor out d𝑦 by dπ‘₯. And finally, we divide to obtain our solution to part one that d𝑦 by dπ‘₯ is equal to π‘₯ over two 𝑦 plus eight.

For part two, we’re required to write an equation of the line tangent to the curve at the point six, zero. Now since the tangent is going to be a straight line, we know that it will be of the form 𝑦 is equal to π‘Žπ‘₯ plus 𝑏, where π‘Ž is the slope of the tangent. And we can find the slope of the tangent using the equation we found in part i since d𝑦 by dπ‘₯ can give the gradient of the curve at any point. Now we need to find the slope of the curve at the point six, zero.

We can simply substitute π‘₯ equals six and 𝑦 equals zero into our equation for d𝑦 by dπ‘₯. In doing this, we obtain that d𝑦 by dπ‘₯ which is also the slope of our tangent is equal to three-fourths. Therefore, we can substitute π‘Ž equals three-fourths into our equation for our tangent. Now all that remains is to find the value of b. We can do this by substituting π‘₯ equals six and 𝑦 equals zero into the equation for our tangent. Rearranging this, we find that 𝑏 is equal to negative nine over two. Now, I put this back into the equation of our tangent to obtain an equation of 𝑦 is equal to three-fourths π‘₯ minus nine over two. This is the equation of the tangent to the curve at the point six, zero.

Part three) Write an equation of each vertical tangent to the curve.

Now, a vertical line has an infinite slope. And so therefore, our vertical tangent will occur when the denominator of d𝑦 by dπ‘₯ is equal to zero since d𝑦 by dπ‘₯ is the equation for the slope of our curve. The denominator of d𝑦 by dπ‘₯ is two 𝑦 plus eight. So we can set that equal to zero. Now, we rearrange this. And we find that 𝑦 is equal to negative four. Now, this is an equation of a line. However, this is a horizontal line. And we have been asked to find the vertical tangents.

In order to find the vertical tangents, we can substitute this value of 𝑦 is equal to negative four back into the original equation. We obtain this which we can simplify and then rearrange to obtain that π‘₯ squared is equal to four. Now, we will take the square root here. But we must be careful to remember that we would obtain both positive and negative solutions. We get π‘₯ is equal to negative two and π‘₯ is equal to two. Now, these are equations of vertical lines. Therefore, π‘₯ equals negative two and π‘₯ equals two are the equations to the vertical tangents of our curve.

Part four) Evaluate d two 𝑦 by dπ‘₯ squared at the point three root six, one.

Now, d two 𝑦 by dπ‘₯ squared is the second differential of 𝑦 with respect to π‘₯ or the differential of d𝑦 by dπ‘₯ with respect to π‘₯. Now, we’ve already found d𝑦 by dπ‘₯ in part one of the question. And so, in order to find d two 𝑦 by dπ‘₯ squared, we simply need to differentiate π‘₯ over two 𝑦 plus eight with respect to π‘₯. Here, we’re differentiating a quotient. And so, we can use the quotient rule. The quotient rule tells us that d by dπ‘₯ of 𝑒 over 𝑣 is equal to 𝑣 d𝑒 by dπ‘₯ minus 𝑒 d𝑣 by dπ‘₯ all over 𝑣 squared.

If we compare this to our equation, we can say that π‘₯ is equal to 𝑒 and two 𝑦 plus eight is equal to 𝑣. Next, we need to find d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. To find d𝑒 by dπ‘₯, we differentiate π‘₯ with respect to π‘₯. This simply gives us one. Now, to find d𝑣 by dπ‘₯ when you differentiate two 𝑦 plus eight with respect to π‘₯, we obtain that d𝑣 by dπ‘₯ is equal to two d𝑦 by dπ‘₯. Now, we can use the quotient rule and substitute in these values. We obtain that d two 𝑦 by dπ‘₯ squared is equal to two 𝑦 plus eight multiplied by one minus π‘₯ multiplied by two d𝑦 by dπ‘₯ all over two 𝑦 plus eight squared.

Now, we’re asked to evaluate d two 𝑦 by dπ‘₯ squared at the point three root six, one. So our next step here would be to substitute in π‘₯ equals three root six and 𝑦 equals one. However, we would not know the value of d𝑦 by dπ‘₯. So let’s start by finding the value of d𝑦 by dπ‘₯ at the point three root six, one. We substitute these values into the equation we found in part one. And then, we can simplify this to obtain that at the point three root six, one, d𝑦 by dπ‘₯ is equal to three root six over 10.

Now, we can evaluate d two 𝑦 by dπ‘₯ squared at the point three root six, one by substituting in the values for π‘₯, 𝑦, and d𝑦 by dπ‘₯. And next, we can simplify this to obtain that at the point three root six, one, d two 𝑦 by dπ‘₯ squared is equal to negative one over 125. This completes the final part of the question.

The solutions we found are as follows. Part one) d𝑦 by dπ‘₯ is equal to π‘₯ over two 𝑦 plus eight. Part two) 𝑦 is equal to three-fourths π‘₯ minus nine over two. Part three) π‘₯ is equal to negative two and π‘₯ is equal to two. And finally, for part four, we evaluated d two 𝑦 by dπ‘₯ squared to be equal to negative one over 125.

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