Question

Please help me with solving these. I’d really appreciate your help. Thank you very much.

Answer 1

Answer:

Problem 17)

$\displaystyle y=-\frac{5}{2}x+\frac{7}{2}$

Problem 18)

$\displaystyle y=\frac{3}{2}x-\frac{3}{2}+\frac{\pi}{4}$

Step-by-step explanation:

Problem 17)

We have the curve represented by the equation:

$\displaystyle 4x^2+2xy+y^2=7$

And we want to find the equation of the tangent line to the point (1, 1).

First, let's find the derivative dy/dx. Take the derivative of both sides with respect to x:

$\displaystyle \frac{d}{dx}\left[4x^2+2xy+y^2\right]=\frac{d}{dx}[7]$

Simplify. Recall that the derivative of a constant is zero.

$\displaystyle \frac{d}{dx}[4x^2]+\frac{d}{dx}[2xy]+\frac{d}{dx}[y^2]=0$

Differentiate. We can differentiate the first term normally. The second term will require the product rule. Hence:

$\displaystyle 8x+\left(2y+2x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0$

Rewrite:

$\displaystyle \frac{dy}{dx}\left(2x+2y\right)=-8x-2y$

Therefore:

$\displaystyle \frac{dy}{dx}=\frac{-8x-2y}{2x+2y}=-\frac{4x+y}{x+y}$

So, the slope of the tangent line at the point (1, 1) is:

$\displaystyle \frac{dy}{dx}\Big|_{(1, 1)}=-\frac{4(1)+(1)}{(1)+(1)}=-\frac{5}{2}$

And since we know that it passes through the point (1, 1), by the point-slope form:

$\displaystyle y-1=-\frac{5}{2}(x-1)$

If desired, we can simplify this into slope-intercept form. Therefore, our equation is:

$\displaystyle y=-\frac{5}{2}x+\frac{7}{2}$

Problem 18)

We have the equation:

$\displaystyle y=\tan^{-1}\left(x^3\right)$

And we want to find the equation of the tangent line to the graph at the point (1, π/4).

Take the derivative of both sides with respect to x:

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\tan^{-1}(x^3)]$

We can use the chain rule:

$\displaystyle \frac{d}{dx}[u(v(x))]=u'(v(x))\cdot v(x)$

Let u(x) = tan⁻¹(x) and let v(x) = x³. Thus:

(Recall that d/dx [arctan(x)] = 1 / (1 + x²).)

$\displaystyle \frac{d}{dx}\left[\tan^{-1}(x^3)\right]=\frac{1}{1+v^2(x)}\cdot 3x^2$

Substitute and simplify. Hence:

$\displaystyle \frac{d}{dx}\left[\tan^{-1}(x^3)\right]=\frac{1}{1+v^2(x)}\cdot 3x^2=\frac{3x^2}{1+x^6}$

Then the slope of the tangent line at the point (1, π/4) is:

$\displaystyle \frac{dy}{dx}\Big|_{x=1}=\frac{3(1)^2}{1+(1)^6}=\frac{3}{2}$

Then by the point-slope form:

$\displaystyle y-\frac{\pi}{4}=\frac{3}{2}(x-1)$

Or in slope-intercept form:

$\displaystyle y=\frac{3}{2}x-\frac{3}{2}+\frac{\pi}{4}$