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We know that upon differentiating a function at a certain point, we can obtain its gradient at that point. That will also be the gradient of its tangent line. Thus,
y = 8x * sin(x)
y' = 8x * cos(x) + 8sin(x)
Evaluating the gradient at the given point, we insert pi/2 for x.
gradient = 8(pi/2) * cos(pi/2) + 8sin(pi/2)
= 2pi*sqrt(2) + 4(sqrt(2))
The equation of the line will be in the form y = mx + c; where m is the gradient and c is the y intercept. To calculate the y-int, we insert the given x and y values:
4pi = [2pi*sqrt(2)+4(sqrt(2))]*(pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98
We know that upon differentiating a function at a certain point, we can obtain its gradient at that point. That will also be the gradient of its tangent line. Thus,
y = 8x * sin(x)
y' = 8x * cos(x) + 8sin(x)
Evaluating the gradient at the given point, we insert pi/2 for x.
gradient = 8(pi/2) * cos(pi/2) + 8sin(pi/2)
= 2pi*sqrt(2) + 4(sqrt(2))
The equation of the line will be in the form y = mx + c; where m is the gradient and c is the y intercept. To calculate the y-int, we insert the given x and y values:
4pi = [2pi*sqrt(2)+4(sqrt(2))]*(pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98
The equation of tangent line is [tex]\boxed{y=8x}[/tex].
Further explanation:
A tangent line is a line which touches a curve at one and only one point.
The equation of a tangent line is [tex]y=mx+c[/tex] where [tex]m[/tex] is the slope of the line is and [tex]c[/tex] is the [tex]y[/tex]-intercept.
Calculation:
The given curve is [tex]y=8x\text{sin}x[/tex].
Differentiate the above curve using product rule of differentiation as follows:
[tex]\begin{aligned}y'&=\dfrac{d}{dx}(8x\text{sin}x)\\&=8\text{sin}x+8x\text{cos}x\end{aligned}[/tex]
Label the above equation as follows:
[tex]\boxed{y'=8\text{sin}x+8x\text{cos}x}[/tex] .......(1)
The point on the curve through which the tangent line passes is [tex]\left(\dfrac{\pi}{2},4\pi\right)[/tex].
Substitute [tex]x=\frac{\pi}{2}[/tex] in equation (1).
[tex]\begin{aligned}y'&=8\text{sin}\left(\dfrac{\pi}{2}\right)+\left(8\cdot \dfrac{\pi}{2}\cdot \text{cos}\left(\dfrac{\pi}{2}\right)\right)\\&=8\cdot 1+\left(8\cdot \dfrac{\pi}{2}\cdot 0\right)\\&=8\end{aligned}[/tex]
Therefore, the slope of the tangent line is [tex]8[/tex].
The equation of the tangent line is [tex]y=mx+c[/tex] where [tex]m[/tex] is the slope and [tex]c[/tex] is the [tex]y[/tex]-intercept.
Now, substitute the value of [tex]y,m\text{ and }x[/tex] in the equation of tangent line to obtain the value of [tex]c[/tex] as follows:
[tex]\begin{aligned}4\pi&=8\cdot \left(\dfrac{\pi}{2}\right)\\ \4\pi&=4\pi +c\\c&=0\end{aligned}[/tex]
The [tex]\bf y[/tex] intercept of the tangent line is [tex]c=0[/tex].
Now, substitute the values of [tex]m[/tex] and [tex]c[/tex] in [tex]y=mx+c[/tex].
[tex]\begin{aligned}y&=8x+0\\y&=8x\end{aligned}[/tex]
Therefore, the equation of tangent is [tex]\boxed{y=8x}[/tex].
Learn more
1. Problem on the equation of the circle https://brainly.com/question/1952668
2. Problem on the center and radius of an equation https://brainly.com/question/9510228
3. Problem on the general form of the equation of the circle https://brainly.com/question/1506955
Answer details:
Grade: High school
Subject: Mathematics
Chapter: Lines and Tangents
Keywords: Tangents, equation of tangent line, slope, intercept, curve, line, differentiation, line, calculus, y=8xsinx, pi/2, y-intercept, derivative, product rule, derivative.
