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

& =\dfrac{dy}{dt}\ \times \ \dfrac{dt}{dx} \\\ & =\dfrac{dy}{dx} \\\ \end{aligned}$$ _Complete step-by-step answer:_ So, as mentioned above, on multiplying the two terms, we will get the value of $$\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{aligned} & 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{aligned}$$ Now, on multiplying $$\dfrac{dy}{dt}$$ with the reciprocal of $$\dfrac{dx}{dt}$$ , we get the following $$\begin{aligned} & \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{aligned}$$ Now, for getting the slope at t=2, we can put the value of t=2 in $$\dfrac{dy}{dx}$$ as follows $$\begin{aligned} & \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{aligned}$$ Hence, the slope of the tangent at t=2 is $$\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)$$ )