Video: Finding the Slope of a Tangent to a Trigonometric Curve

The graph of 𝑦 = βˆ’1 βˆ’ (2 sin π‘₯) has a zero on the interval [πœ‹/2, 3πœ‹/2]. What is the slope of the tangent to the graph at that point?

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Video Transcript

The graph of 𝑦 equals negative one minus two sin π‘₯ has a zero on the closed interval πœ‹ by two to three πœ‹ by two. What is the slope of the tangent to the graph at that point?

Well firstly, we know that, to find the slope of the tangent to a graph when given a function 𝑓 of π‘₯, we need to evaluate the derivative at that point. The problem is, we don’t know where that point is yet. So we use the fact that the graph of 𝑦 equals negative one minus two sin of π‘₯ has a zero on the closed interval between πœ‹ by two and three πœ‹ by two radians. πœ‹ has a zero. This tells us that the graph intersects the π‘₯-axis in this interval.

And so to find the value of π‘₯ at this point, we let 𝑦 be equal to zero. So we get negative one minus two sin of π‘₯ equals zero. We need to solve for π‘₯. To achieve this, we add two sin π‘₯ to both sides of our equation and then divide by two, so negative one-half equals sin of π‘₯. Then we take the inverse or arcsin of both sides of our equation. So we find π‘₯ is equal to the inverse sin of negative one-half.

Well, we know this is true for one value; specifically, that’s πœ‹ by six. The problem is, that’s outside of our closed interval. However, we know that, for the sine curve, there is another solution. And we find that by subtracting πœ‹ by six from πœ‹. And we can find that value either by sketching the graph of 𝑦 is equal to sin π‘₯ and looking for various points of symmetry or by using the CAST diagram.

Now we need to find the slope of the tangent to the graph at the point where π‘₯ equals five πœ‹ by six. Well, the slope is found by differentiating 𝑦 with respect to π‘₯. We differentiate each term in turn. The derivative of negative one is zero. And when we differentiate negative two sin π‘₯, we get negative two cos π‘₯. Let’s evaluate this at the point where π‘₯ equals five πœ‹ by six. So that’s negative two times cos of five πœ‹ by six, which is equal to the square root of three.

And so the slope of the tangent to the graph of 𝑦 equals negative one minus two sin π‘₯ at the point where π‘₯ equals five πœ‹ by six is root three.

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