Question

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

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

Answer 1

Answer: (B)

$\frac{6}{7}$

Differentiating $x,y$ w.r.t $t,$ we get

$\frac{dy}{dt}=\frac{d}{dt}(2t^2-2t-5)=4t-2\frac{dx}{dt}=\frac{d}{dt}(t^2+3t-8)=2t+3$

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

At $t=2$ derivative will be $\frac{dy}{dx}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}$

Thus, the slope of tangent is $\frac{6}{7}$.