To find the slope of the tangent line to the curve defined by the equation 4x^2+5xy+y^4 =370 at the point (-9,-1), we need to use implicit differentiation.
Differentiating both sides with respect to x, we get:
d/dx (4x^2+5xy+y^4) = d/dx (370)
8x + 5y + 4y^3 (dy/dx) = 0 [using the chain rule on y^4]
We can find the value of dy/dx at the point (-9,-1) by substituting these values into the equation:
8(-9) + 5(-1) + 4(-1)^3 (dy/dx) = 0
-72 - 5 + 4(dy/dx) = 0
dy/dx = (72 + 5)/4
dy/dx = 77/4
Therefore, the slope of the tangent line to the curve at the point (-9,-1) is 77/4.
First of all, we must know that the slope of a straight line is nothing more than the angle "θ" that the referred straight line forms with the axis of the abscissas in its positive direction, and to determine its value we must obtain the measure of the arc whose tangent is equal to the angularcoefficient, that is:
Furthermore, we must realize that the slope of the tangent line is numerically equal to the first derivative of the function at the said point of tangency, that is:
Note now that the equation of the curve is given implicitly. In this way, the derivative of the curve must be calculatedimplicitly in relation to the variable "x". So, we do:
[tex]\:\:\Large\displaystyle\text{$\begin{gathered} \sf t: y = -\frac{77}{49}x - \frac{742}{49}\end{gathered}$}[/tex]
[tex]\\\\[/tex]
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10 votes Thanks 9
ItsZen
Kasi sa law of integers if 2 negative sign is equivalent to positive, wag mona rin siyang i construct kasi sabi sa question, just defined lang po then yun na 77/49 yung sagot.
ItsZen
I don't know if my explanation is correct hahà
Deleted account
Actually, -77/49x - 742/49 is the final answer, further contructed by -77x-742/49. I even recalculate the problem, due to your complaint HAHA
Answers & Comments
To find the slope of the tangent line to the curve defined by the equation 4x^2+5xy+y^4 =370 at the point (-9,-1), we need to use implicit differentiation.
Differentiating both sides with respect to x, we get:
d/dx (4x^2+5xy+y^4) = d/dx (370)
8x + 5y + 4y^3 (dy/dx) = 0 [using the chain rule on y^4]
We can find the value of dy/dx at the point (-9,-1) by substituting these values into the equation:
8(-9) + 5(-1) + 4(-1)^3 (dy/dx) = 0
-72 - 5 + 4(dy/dx) = 0
dy/dx = (72 + 5)/4
dy/dx = 77/4
Therefore, the slope of the tangent line to the curve at the point (-9,-1) is 77/4.
I hope this helps
Verified answer
After solving the calculations, we conclude that the slope of the tangent line to the curve given by the point of tangency "T(-9, -1)" is:
[tex]\qquad\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bold{\theta \cong 122.47^{\circ}\:\:\:}}}\end{gathered}$}[/tex]
Let the data be:
First of all, we must know that the slope of a straight line is nothing more than the angle "θ" that the referred straight line forms with the axis of the abscissas in its positive direction, and to determine its value we must obtain the measure of the arc whose tangent is equal to the angular coefficient, that is:
[tex]\Large\displaystyle\text{$\begin{gathered}{\bold{ I}\:\:\:\:\qquad \theta = \arctan(m_{ t})}\end{gathered}$}[/tex]
Furthermore, we must realize that the slope of the tangent line is numerically equal to the first derivative of the function at the said point of tangency, that is:
[tex]\Large\displaystyle\text{$\begin{gathered} {\bold{ II}\:\:\:\:\:\:\:\:\:\:\:m_{r} = f'(x_{ T})}\end{gathered}$}[/tex]
Replacing "II" in "I", we have:
[tex]\Large\displaystyle\text{$\begin{gathered} {\bold{ III}\:\:\:\:\: \theta = \arctan\left[f'(x_{T})\right]}\end{gathered}$}[/tex]
Note now that the equation of the curve is given implicitly. In this way, the derivative of the curve must be calculated implicitly in relation to the variable "x". So, we do:
[tex]\begin{gathered}\large \text {$\begin{aligned} \sf (4x^2 + 5xy + y^4)' & \sf \: = (370)'\\ \sf 2\cdot4\cdot x^{2 - 1}+ 1 \cdot 5\cdot x^{1 - 1}\cdot y + 5xy' + 4\cdot y^{4 -1}\cdot y' & \sf \: = 0\\ \sf 8x + 5y + 5xy' + 4y^3y' & \sf \: = 0\\ \sf 5xy' + 4y^3y' & \sf \: = -8x - 5y\\ \sf (5x + 4y^3)y' & \sf \: = -8x - 5y\\ \sf y' & \sf \: = \frac{-8x - 5y}{5x + 4y^3}\end{aligned} $}\end{gathered}[/tex]
Therefore, the implicit derivative of the curve with respect to the variable "x" is:
Knowing that:
So we can rewrite the derivative as:
Substituting the value of the derivative in equation "III", we have:
[tex]\Large\displaystyle\text{$\begin{gathered} {\bold{ IV}\:\:\:\:\:\:\:\theta = \arctan\left[\frac{-8x - 5y}{5x + 4y^3}\right]}\end{gathered}$}[/tex]
Substituting the coordinates of the point of tangency in the "IV" equation, we have:
[tex]\begin{gathered}\Large \text {$\begin{aligned} \sf\theta & \sf \: = \arctan\left[\frac{-8\cdot(-9) - 5\cdot(-1)}{5\cdot( -9) + 4\cdot(-1)^3}\right]\\& \sf \: = \arctan\left[\frac{72 + 5}{-45 - 4}\right]\\& \sf \: = \arctan\left [-\frac{77}{49}\right]\\& \: \sf{\cong 122.47^{\circ}}\end{aligned} $}\end{gathered}[/tex]
Therefore, the slope of the sought tangent line is:
From this, we can construct the equation of the tangent line. For that, we do:
[tex]\begin{gathered}\Large \text {$\begin{aligned} \sf y - y_{T} & \sf \: = f'(x_{T})\cdot(x - x_{T}) \\ \sf y - (-1) & \sf \: = -\frac{77}{49}\cdot(x - (-9))\\ \sf y + 1 & \sf \: = -\frac{77}{49}\cdot(x + 9)\\ \sf y + 1 & \sf \: = -\frac{77}{49}x - \frac{693}{49}\\ \sf y & \sf \: = -\frac{77}{49}x - \frac{693}{49} -1\\ \sf y & \sf \: = \frac{-77x - 693 - 49}{49} \\ \sf y & \sf \: = \frac{-77x - 742}{49}\\ \sf y & \sf \: = -\frac{77}{49}x - \frac{742 }{49}\end{aligned} $}\end{gathered}[/tex]
Therefore, the tangent line is:
[tex]\\\\[/tex]
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