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12. What is the equation of the line tangent to the graph of y=sin⁡2xy=\sin ^{2} x at...
Aug 13, 2024
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Solution by Steps
step 2
Next, we find the derivative of y=sin⁡2xy = \sin^2 x using the chain rule: dydx=2sin⁡xcos⁡x=sin⁡(2x)\frac{dy}{dx} = 2 \sin x \cos x = \sin(2x). Evaluating this at x=π4x = \frac{\pi}{4} gives dydx=sin⁡(π2)=1\frac{dy}{dx} = \sin\left(\frac{\pi}{2}\right) = 1
step 3
The slope of the tangent line at the point (π4,12)\left(\frac{\pi}{4}, \frac{1}{2}\right) is 1. We can use the point-slope form of the equation of a line: y−y1=m(x−x1)y - y_1 = m(x - x_1), where (x1,y1)=(π4,12)(x_1, y_1) = \left(\frac{\pi}{4}, \frac{1}{2}\right) and m=1m = 1
step 4
Substituting these values into the point-slope form gives us: y−12=1(x−π4)y - \frac{1}{2} = 1\left(x - \frac{\pi}{4}\right), which simplifies to y−12=x−π4y - \frac{1}{2} = x - \frac{\pi}{4}
step 5
Therefore, the equation of the tangent line is y−12=(x−π4)y - \frac{1}{2} = (x - \frac{\pi}{4})
B
Key Concept
Tangent Line Equation
Explanation
The equation of the tangent line can be found using the point-slope form, which requires the slope at the point of tangency and the coordinates of that point.
What is the derivative of the function y=sin⁡2x y = \sin^2 x , and how would you use it to find the slope of the tangent line at x=π4 x = \frac{\pi}{4} ?
Generated Graph
Solution by Steps
step 1
To find the derivative of the function y=sin⁡2x y = \sin^2 x , we use the chain rule: dydx=2sin⁡xcos⁡x=sin⁡(2x) \frac{dy}{dx} = 2 \sin x \cos x = \sin(2x)
step 2
Next, we evaluate the derivative at x=π4 x = \frac{\pi}{4} : dydx∣x=π4=sin⁡(2⋅π4)=sin⁡(π2)=1 \frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1
step 3
The slope of the tangent line at x=π4 x = \frac{\pi}{4} is therefore 1 1
step 4
Now, we find the point on the curve at x=π4 x = \frac{\pi}{4} : y=sin⁡2(π4)=12 y = \sin^2\left(\frac{\pi}{4}\right) = \frac{1}{2}
step 5
Using the point-slope form of the line, the equation of the tangent line is y−12=1(x−π4) y - \frac{1}{2} = 1\left(x - \frac{\pi}{4}\right) , which simplifies to y−12=x−π4 y - \frac{1}{2} = x - \frac{\pi}{4}
Answer
The equation of the tangent line is y−12=(x−π4) y - \frac{1}{2} = (x - \frac{\pi}{4}) , which corresponds to answer choice B.
Key Concept
The derivative of a function gives the slope of the tangent line at any point on the graph.
Explanation
By finding the derivative and evaluating it at a specific point, we can determine the slope of the tangent line, which is essential for writing the equation of that line.
What is the second derivative of the function y=sin⁡2x y = \sin^2 x ?
Generated Graph
Solution by Steps
step 1
To find the second derivative of the function y=sin⁡2x y = \sin^2 x , we first compute the first derivative: dydx=2sin⁡xcos⁡x=sin⁡(2x) \frac{dy}{dx} = 2 \sin x \cos x = \sin(2x)
step 2
Next, we differentiate sin⁡(2x) \sin(2x) to find the second derivative: d2ydx2=2cos⁡(2x) \frac{d^2y}{dx^2} = 2 \cos(2x)
step 3
We can also express this as 2cos⁡2x−2sin⁡2x=2cos⁡(2x) 2 \cos^2 x - 2 \sin^2 x = 2 \cos(2x)
step 4
Therefore, the second derivative can be simplified to 2−4sin⁡2x 2 - 4 \sin^2 x
Answer
The second derivative of y=sin⁡2x y = \sin^2 x is 2cos⁡(2x) 2 \cos(2x) or 2−4sin⁡2x 2 - 4 \sin^2 x .
Key Concept
The second derivative measures the curvature of the function, indicating how the slope of the tangent line changes.
Explanation
The second derivative provides information about the concavity of the function, which is essential for understanding its behavior.
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