Find an equation of the tangent line to the curve y=8x sinx at the point pi/2 4pi

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hakeya19013 hakeya19013

27-09-2016
Mathematics

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Find an equation of the tangent line to the curve y=8x sinx at the point pi/2 4pi

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scme1702 scme1702 · Answer 1 · 2016-10-06T14:31:08-04:00

Hello
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)
= 2pisqrt(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 = [2pisqrt(2)+4(sqrt(2))](pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98

AkhileshT AkhileshT · Answer 2 · 2019-09-07T03:10:52-04:00

The equation of tangent line is \(\boxed{y=8x}\).

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 \(y=mx+c\) where \(m\) is the slope of the line is and \(c\) is the \(y\)-intercept.

Calculation:

The given curve is \(y=8x\text{sin}x\).

Differentiate the above curve using product rule of differentiation as follows:

\(\begin{aligned}y'&=\dfrac{d}{dx}(8x\text{sin}x)\\&=8\text{sin}x+8x\text{cos}x\end{aligned}\)

Label the above equation as follows:

\(\boxed{y'=8\text{sin}x+8x\text{cos}x}\) .......(1)

The point on the curve through which the tangent line passes is \(\left(\dfrac{\pi}{2},4\pi\right)\).

Substitute \(x=\frac{\pi}{2}\) in equation (1).

\(\begin{aligned}y'&=8\text{sin}\left(\dfrac{\pi}{2}\right)+\left(8\cdot \dfrac{\pi}{2}\cdot \text{cos}\left(\dfrac{\pi}{2}\right)\right)\\&=8\cdot 1+\left(8\cdot \dfrac{\pi}{2}\cdot 0\right)\\&=8\end{aligned}\)

Therefore, the slope of the tangent line is \(8\).

The equation of the tangent line is \(y=mx+c\) where \(m\) is the slope and \(c\) is the \(y\)-intercept.

Now, substitute the value of \(y,m\text{ and }x\) in the equation of tangent line to obtain the value of \(c\) as follows:

\(\begin{aligned}4\pi&=8\cdot \left(\dfrac{\pi}{2}\right)\\ \4\pi&=4\pi +c\\c&=0\end{aligned}\)

The \(\bf y\) intercept of the tangent line is \(c=0\).

Now, substitute the values of \(m\) and \(c\) in \(y=mx+c\).

\(\begin{aligned}y&=8x+0\\y&=8x\end{aligned}\)

Therefore, the equation of tangent is \(\boxed{y=8x}\).

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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.