Question

What is the slope to the tangent to the curve \[x={{t}^{2}}+3t-8\], \[y=2{{t}^{2}}-2t-5\] at t=2 ?
A) \[\dfrac{7}{6}\]
B) \[\dfrac{6}{7}\]
C) 1
D) \[\dfrac{5}{6}\]

Answer 1

HINT: Now, in this question, we will first take the derivatives of x (which is a function of t) and y (which is a function of t) with respect to t that means we will get the values of the following
\[\dfrac{dy}{dt}\ and\ \dfrac{dx}{dt}\]
Now, we will take the reciprocal of \[\dfrac{dx}{dt}\] and then we will multiply it with \[\dfrac{dy}{dt}\] which would be as follows
\[\begin{align}
  & =\dfrac{dy}{dt}\ \times \ \dfrac{dt}{dx} \\
 & =\dfrac{dy}{dx} \\
\end{align}\]

Complete step-by-step answer:
So, as mentioned above, on multiplying the two terms, we will get the value of \[\dfrac{dy}{dx}\] (which is the slope of the function of which x and y coordinates behave as it is given in the question)
Now, we will put the value of t=2 in \[\dfrac{dy}{dx}\] and hence, we will get the slope.
As mentioned in the question, we have to find the slope of the function at t=2.
Now, for that we can simply follow what is given in the hint as follows
First we will calculate \[\dfrac{dy}{dt}\ and\ \dfrac{dx}{dt}\] as follows
\[\begin{align}
  & y=2{{t}^{2}}-2t-5 \\
 & \dfrac{dy}{dt}=4t-2 \\
 & \left( \dfrac{d\left( {{x}^{n}} \right)}{dx} \right)=n{{x}^{n-1}}\ and\,\left( \dfrac{d\left(\text{constant }\right)}{dx} \right)=0 \\
 & \\
 & x={{t}^{2}}+3t-8 \\
 & \dfrac{dx}{dt}=2t+3 \\
 & \left( \dfrac{d\left( {{x}^{n}} \right)}{dx} \right)=n{{x}^{n-1}}\ and\,\left( \dfrac{d\left(\text{constant }\right)}{dx} \right)=0 \\
\end{align}\]
Now, on multiplying \[\dfrac{dy}{dt}\] with the reciprocal of \[\dfrac{dx}{dt}\] , we get the following
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dt}\times \dfrac{dt}{dx}=\dfrac{dy}{dx} \\
 & \Rightarrow \left( 4t-2 \right)\times \dfrac{1}{\left( 2t+3 \right)}=\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{\left( 4t-2 \right)}{\left( 2t+3 \right)}=\dfrac{dy}{dx} \\
\end{align}\]
Now, for getting the slope at t=2, we can put the value of t=2 in \[\dfrac{dy}{dx}\] as follows
\[\begin{align}
  & \Rightarrow \dfrac{\left( 4t-2 \right)}{\left( 2t+3 \right)}=\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{\left( 4\times 2-2 \right)}{\left( 2\times 2+3 \right)}=\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{\left( 8-2 \right)}{\left( 4+3 \right)}=\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{\left( 6 \right)}{\left( 7 \right)}=\dfrac{dy}{dx} \\
\end{align}\]
Hence, the slope of the tangent at t=2 is \[\dfrac{6}{7}\] .

NOTE:- Now, if we are asked to get the equation of the tangent, we will do the following
We will get the values of x and y at t=2 by simply putting the value of t in x and y and then we will write as follows
\[{{\left. \dfrac{dy}{dx} \right|}_{t=2}}=\dfrac{y-y(2)}{x-x(2)}\]
(Because, the tangent to a curve at any particular value of the variable is \[{{\left. \dfrac{dy}{dx} \right|}_{x={{x}_{1}}\ and\ y={{y}_{1}}}}=\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}\] at\[\left( {{x}_{1}},{{y}_{1}} \right)\] )