Video: Finding the Equation of a Curve given Its Second Derivative and the Slope of That Tangent at a Point on the Curve

The second derivative of a curve is βˆ’27 sin 3π‘₯ + 8. The curve passes through the point (πœ‹/6, (βˆ’4πœ‹/3) + (πœ‹Β²/9) + 6) and the gradient of the tangent at this point is βˆ’8 + (4πœ‹/3). Find the equation of the curve.

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Video Transcript

The second derivative of a curve is negative 27 sin three π‘₯ plus eight. The curve passes through the point πœ‹ by six, negative four πœ‹ by three plus πœ‹ squared over nine plus six. And the gradient of the tangent at this point is negative eight plus four πœ‹ over three. Find the equation of the curve.

The first piece of information we’re given here is the expression for the second derivative of the curve. That’s d two 𝑦 by dπ‘₯ squared, and it’s equal to negative 27 sin of three π‘₯ plus eight. We’re given some information about a point that this curve passes through, though we can’t use that just yet. And we also know the gradient of the tangent to the curve at this point.

So we should begin by finding a relationship between the second derivative, the gradient of the tangent, and an equation for the curve. Well, we should begin by recalling that integration and differentiation are essentially reverse processes of one another. And so if we were to integrate the expression for the second derivative with respect to π‘₯, we form an expression for the first derivative. And, of course, the first derivative tells us how to find the gradient or the slope of the tangent to the curve at any point.

So let’s integrate negative 27 sin of three π‘₯ plus eight with respect to π‘₯. We can quote the general result for the integral of sin of π‘Žπ‘₯ with respect to π‘₯ for real constant values of π‘Ž. It’s negative one over π‘Ž cos of π‘Žπ‘₯ plus some constant of integration 𝐢. And so the integral of negative 27 sin of three π‘₯ is negative 27 times negative one-third cos of three π‘₯.

Then when we integrate eight, we get eight π‘₯. So d𝑦 by dπ‘₯ is equal to negative 27 times negative one-third cos of three π‘₯ plus eight π‘₯ plus some constant of integration 𝐢. We need to find the value of 𝐢. And so we’re going to use the information that the curve passes through the point πœ‹ by six, negative four πœ‹ by three plus πœ‹ squared over nine plus six. And the gradient of the tangent at this point is negative eight plus four πœ‹ by three.

This really tells us that when π‘₯ is equal to πœ‹ by six, d𝑦 by dπ‘₯ is equal to negative eight plus four πœ‹ by three. So let’s replace d𝑦 by dπ‘₯ and π‘₯ in our equation. We get negative eight plus four πœ‹ over three equals negative 27 times negative one-third cos of three times πœ‹ by six plus eight times πœ‹ by six plus 𝐢.

Simplifying the right-hand side, and we get negative 27 times negative one-third cos of πœ‹ by two plus four πœ‹ by three plus 𝐢. We’ll subtract four πœ‹ by three from both sides. And cos of πœ‹ by two is zero. So we see 𝐢 is equal to negative eight. We simplify negative 27 times negative a third cos of three π‘₯ to get positive nine cos of three π‘₯. And we find d𝑦 by dπ‘₯ is nine cos of three π‘₯ plus eight π‘₯ minus eight.

We’re not quite finished though. We’re trying to find the equation of the curve. Once again, we recall the relationship between integration and differentiation. And we see that 𝑦 is equal to the integral of d𝑦 by dπ‘₯ with respect to π‘₯. That’s the integral of nine cos three π‘₯ plus eight π‘₯ minus eight dπ‘₯. This time, we quote the general result for the integral of cos of π‘Žπ‘₯. It’s one over π‘Ž sin of π‘Žπ‘₯.

And so the integral of nine times cos of three π‘₯ is nine times a third sin of three π‘₯. Then the integral of eight π‘₯ is eight π‘₯ squared over two, and the integral of negative eight is negative eight π‘₯. So 𝑦 is nine times a third sin of three π‘₯ plus eight π‘₯ squared over two minus eight π‘₯ plus a constant of integration 𝐷. This simplifies to 𝑦 equals three sin of three π‘₯ plus four π‘₯ squared minus eight π‘₯ plus 𝐷.

And this time, we can find the value of 𝐷 by going back to this information. When π‘₯ is πœ‹ by six, 𝑦 is negative four πœ‹ over three plus πœ‹ squared over nine plus six.

Let’s clear some space and complete this final step. When we do, we see that negative four πœ‹ over three plus πœ‹ squared over nine plus six equals three sin of three times πœ‹ over six plus four times πœ‹ over six squared minus eight times πœ‹ over six plus 𝐷. The right-hand side here simplifies to three plus πœ‹ squared over nine minus four πœ‹ over three plus 𝐷. Then we see that we can add four πœ‹ over three to both sides and subtract πœ‹ squared over nine. Finally, we subtract three from both sides, and we find that 𝐷 is six minus three, which is three.

And so we’ve found the equation of the curve to be 𝑦 equals three sin of three π‘₯ plus four π‘₯ squared minus eight π‘₯ plus three. Which we might alternatively choose to write as 𝑦 equals four π‘₯ squared minus eight π‘₯ plus three sin three π‘₯ plus three.

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