Question

What is the slope of the tangent to the curve$\mathrm{x}={\mathrm{t}}^2+3\mathrm{t}-8,\mathrm{y}=2{\mathrm{t}}^2-2\mathrm{t}-5$ at $\mathrm{t}=2$?

• A. $\frac{7}{6}$
• B. $\frac{6}{7}$
• C. $1$
• D. $\frac{5}{6}$

Answer 1

The correct option is B

$\frac{6}{7}$

Explanation for the correct option:

Step 1: Find the derivative of $x$ and $y$ with respect to $t$.

The slope of a tangent to the curve is determined by finding the derivative.

$\mathrm{x}={\mathrm{t}}^2+3\mathrm{t}-8$

Differentiate with respect to $t$

$\frac{dx}{dt}=\frac{d}{dt}·t^2+3\frac{d}{dt}·t-0$ $\left[\because \frac{d}{dt}\left(cons\tan t\right)=0\right]$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=2\mathrm{t}+3$ $\left[\because \frac{d}{dt}\left(k{t}^n\right)=kn{t}^{n-1}\right]$

Now, $\mathrm{y}=2{\mathrm{t}}^2-2\mathrm{t}-5$

Differentiate with respect to $t$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}·2{\mathrm{t}}^2-2\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}-0$ $\left[\because \frac{d}{dt}\left(cons\tan t\right)=0\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=4\mathrm{t}-2$ $\left[\because \frac{d}{dt}\left(k{t}^n\right)=kn{t}^{n-1}\right]$

We know that,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\displaystyle \frac{\mathrm{dy}}{\mathrm{dt}}}}{{\displaystyle \frac{\mathrm{dx}}{\mathrm{dt}}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{t}-2}{2\mathrm{t}+3}$

Step 2: Find the slope of the tangent at $\mathrm{t}=2$

Put the value of $\mathrm{t}=2$ to find the slope.

$\frac{dy}{dx}=\frac{4\left(2\right)-2}{2\left(2\right)+3}$

$\Rightarrow \frac{dy}{dx}=\frac{6}{7}$

The slope of the tangent to the curve $\mathrm{x}={\mathrm{t}}^2+3\mathrm{t}-8,\mathrm{y}=2{\mathrm{t}}^2-2\mathrm{t}-5$ at $\mathrm{t}=2$ is $\frac{6}{7}$.

Hence, the correct answer is option (B).