题目

Problem

The curve $C$ has equation

3x+5y^2+4x^2y=10(2^x)+35\qquad y>0

(a) Find an expression for $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

(6)

Curve $C$ cuts the $y$ -axis at the point $P$ .

(b) Find the exact value of the gradient of the tangent to $C$ at $P$ .

(2)

题目中文翻译

曲线 $C$ 的方程为

3x+5y^2+4x^2y=10(2^x)+35\qquad y>0

(a) 用 $x,y$ 表示 $\dfrac{dy}{dx}$ 。

曲线 $C$ 与 $y$ 轴交于点 $P$ 。

(b) 求曲线 $C$ 在点 $P$ 处切线斜率的精确值。

解答

We have

3x+5y^2+4x^2y=10(2^x)+35

Differentiate implicitly with respect to $x$ .

Term by term,

\frac{d}{dx}(3x)=3

and

\frac{d}{dx}(5y^2)=10y\frac{dy}{dx}

For $4x^2y$ , use the product rule:

\frac{d}{dx}(4x^2y) =8xy+4x^2\frac{dy}{dx}

Also,

\frac{d}{dx}\left(10(2^x)\right)=10(2^x\ln2)

Therefore

3+10y\frac{dy}{dx}+8xy+4x^2\frac{dy}{dx} =10(2^x\ln2)

Collect the terms involving $\dfrac{dy}{dx}$ :

\left(10y+4x^2\right)\frac{dy}{dx} =10(2^x\ln2)-3-8xy

Hence

\boxed{ \frac{dy}{dx} =\frac{10(2^x\ln2)-3-8xy}{10y+4x^2} }

Point $P$ is on the $y$ -axis, so at $P$ ,

x=0

Substitute $x=0$ into the curve:

3(0)+5y^2+4(0)^2y=10(2^0)+35

5y^2=45

Thus

y^2=9

Since $y>0$ ,

y=3

Now substitute $x=0,\ y=3$ into the expression for $\dfrac{dy}{dx}$ :

\begin{aligned} \frac{dy}{dx} &=\frac{10(2^0\ln2)-3-8(0)(3)}{10(3)+4(0)^2} \\ &=\frac{10\ln2-3}{30} \end{aligned}

Therefore the exact gradient of the tangent at $P$ is

\boxed{\frac{10\ln2-3}{30}}